Strong Acid–Strong
Base Mixture
Calculations
Example 2
Here we’ll go over an example in which a strong acid is mixed with a strong base, and we
calculate the pH of the final mixture.
125.0 mL of 0.150 M H2SO4 is mixed
with 150.0 mL of 0.200 M KOH.
We’re given that 125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M
KOH.
125.0 mL of 0.150 M H2SO4 is mixed
with 150.0 mL of 0.200 M KOH. What
is the pH of the final mixture?
And we’re asked to determine the pH of the final mixture.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Just a few words about sulphuric acid, H2SO4.
H2SO4
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
H2SO4.
H2SO4 is a Diprotic Acid
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
is a Diprotic Acid
H2SO4 is a Diprotic Acid
It has 2 protons it can lose.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Which means it has 2 protons it can lose.
When H2SO4 is added to water, it ionizes completely to
lose its first proton:
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
As soon as H2SO4 is added to water, it ionizes completely to lose its first proton:
When H2SO4 is added to water, it ionizes completely to
lose its first proton:
H 2SO4  H 2O  H 3O


 HSO4
 100% ionization 
H+
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
100% of the H2SO4 molecules lose one proton (click) to form hydronium and hydrogen
sulphate ions.
When H2SO4 is added to water, it ionizes completely to
lose its first proton:
H 2SO4  H 2O  H 3O


 HSO4
 100% ionization 
But its second proton does not come off as easily in
water:
HSO4  H 2O
Weak
Acid
H 3O   SO42   100% ionization 
Equilibrium
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
But when its just in water, the second proton does not come off as easily. This proton comes off
when HSO4 minus ionizes. But HSO4- is a weak acid so its ionization in water is very limited.
When H2SO4 is mixed with the
strong base KOH, this is a totally
different situation.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
However, when H2SO4 is mixed with the STRONG BASE KOH, this is a totally
different situation.
H+
H 2SO4  H 2O  H 3O  HSO4
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
When an H2SO4 molecule enters water, it loses one proton (click) to water, to form a
hydronium ion (H3O+) and a hydrogen sulphate ion (HSO4 minus).
H
H+
O
H
O
O
S
O H
O
–
H+
H 2SO4  H 2O  H 3O  HSO4
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Models of these are shown here. Take a moment to check the atoms and the charges and
see how the formulas relate to the structural models.
H
H+
O
H
O
O
S
O H
O
–
2KOH  2K+ + 2OH–
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
When the strong base KOH dissociates in water it forms K+ and OH minus ions. Here we
doubled everything in the equation.
H
H+
H
O
– O
H
O
O
S
O
O
H
O
H
–
–
2KOH  2K+ + 2OH–
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
we show models of the two hydroxide ions from the KOH
H
H+
H
O
– O
H
O
O
S
O
O
H
O
H
–
–
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
One of the hydroxide ions collides with the hydronium ion
H
H
O
H
O
H
O
O
S
O
O
H
O
H
–
–
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
and takes away a proton, to form 2 water molecules
H
H
O
H
O
H
O
O
S
O
O
H
O
H
–
–
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
The other hydroxide ion collides with the hydrogen sulphate ion (click)
H
H
O
H
O
H
O
O
S
O H OH
O
–
–
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
And takes a proton from it to form a water molecule and a suphate ion
H
H
O
H
O
H
H
O
O
S
O
O
H
–
O
–
Sulphate SO42–
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
the sulphate ion has the formula SO4 2 minus.
Two OH– ions were able to remove
both protons from H2SO4.
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
So, in an indirect way, 2 hydroxide ions are able to remove both protons from a
molecule of H2SO4. We‘ll show this with equations.
Two OH– ions were able to remove
both protons from H2SO4.
H 2SO4  H 2O  H 3O


 HSO4
Step 1
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
As soon as H2SO4 is added to water it ionizes completely to form a hydronium ion and a
hydrogen sulphate ion. We’ll call this Step 1
Two OH– ions were able to remove
both protons from H2SO4.

H 2SO4  H 2O  H 3O 
+
OH

HSO4

2H 2O
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
When we add a strong base, one OH minus ion neutralizes the hydronium ion to form 2
water molecules. We’ll call this step 2
Two OH– ions were able to remove
both protons from H2SO4.

H 2SO4  H 2O  H 3O 
+
OH

2H 2O

HSO4
+
OH

2
H 2O  SO4
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
And the other OH minus ion reacts with hydrogen sulphate to form water and a sulphate
ion. We’ll call this Step 3.
Even though
we know that
these steps
occur…
1
2
3
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Even though we know these 3 steps occur when we add H2SO4 to water and then add a
strong base,
Even though
we know that
these steps
occur…
1
2
3
We can represent the process
with a net overall equation.

H2SO4  2OH
2
 2H2O  SO4
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
We can represent the process with a net overall equation: H2SO4 plus 2 OH minus
form 2H2O plus SO4 2minus.
Each H2SO4
donates
2 protons to the
OH– ions
2 H+

H 2SO4  2OH 
2
2H 2O  SO4
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
so in the overall net reaction, we see that each H2SO4 (click) donates 2 protons or H+
ions to the hydroxide ions.
Each H2SO4
donates
2 protons to the
OH– ions

2 mol H
1 mol H 2SO4
2 H+

H 2SO4  2OH 
2
2H 2O  SO4
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
From this, we can write the conversion factor stating there are 2 moles of H+ per 1 mole of H2SO4.
We can use this conversion factor in any calculation where H2SO4 reacts with a strong base.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O  
2 mol H 
4
 1mol
logH2SO
0.0273

  1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Now we’ll do the calculations for this problem. We’ll begin by calculating the initial
moles of H+ added.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1 mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O  
2 mol H 
4
 1mol
logH2SO
0.0273

  1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Its equal to 0.150 moles of H2SO4 per Litre…
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O  
2 mol H 
4
 1mol
logH2SO
0.0273

  1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Times 2 moles of H+ to 1 mole of H2SO4…
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Times 0.125 L
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Which comes to 0.0375 moles of H+. Notice moles of H2SO4 and Litres cancel out.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.03003mol OH 
significant
1L
1mol KOH


figures
Excess mol H  0.0375 mol H  0.0300 mol OH  0.0075 mol H 
+


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
In order to preserve 3 significant figures (the lowest number of significant figures in the
given data)…
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
4 decimal
0.200 mol KOH 1mol OH 



 0.150 L  0.0300
mol
OH
places
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
The answer to this must be expressed to 4 decimal places.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Now we’ll calculate the initial moles of OH minus added.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
It is equal to 0.200 moles of KOH per L
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Times 1 mole of OH minus to 1 mole of KOH
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Times 0.150 L
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Which comes out to 0.0300 moles of OH minus. You can see that moles of KOH and Litres both
cancel. Notice we also have 3 significant figures and 4 decimal places in this answer.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Now we compare the initial moles of H+ and OH minus. We see that 0.0375, the moles of
H+, is greater than 0.0300, the moles of OH minus
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
In Excess
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Limiting
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075
mol H 
Reagent


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
So the H+ is in excess and the OH minus is the limiting reagent.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
We calculate the excess moles of H+…
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
By taking 0.0375 moles of H+
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
and subtracting 0.0300 moles of OH minus.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
To give us 0.0075 moles of H+ in excess
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
4 decimal
4 decimal
 H 3O+    H + 4decimal

 0.0273 M

   places
L
0.275 L places
 0.125 L  0.150places
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
This answer, 0.0075, has 4 decimal places, because the numbers we subtracted both had
4 decimal places.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273
M
2

    0.125 L  0.150 L  
0.275 L significant
pH   log  H 3O    log  0.0273   1.56

figures
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
But we can see that written this way, this number has only 2 significant figures, the 7 and the
5. Therefore the final answer to this problem cannot have more than 2 significant figures.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
The next step on the way to pH, is to find the hydronium ion concentration.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Which is equal to the concentration of H+. These are synonymous in chemistry dealing
with aqueous solutions.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
The concentration of H+ is equal to moles of H+ per Litre of solution. The moles of H+ is
0.0075 moles.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
And the total volume of the mixture is 0.125 L of H2SO4 ….
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Plus 0.150 L of KOH.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
So the concentration of H+ or H3O+ is 0.0075 moles over 0.275 L.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Which comes out to 0.0273 molar. We’ll carry one more significant figure than the 2 our
final answer is limited to. We’ll round to 2 significant figures at the end.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
In the last step, we’ll find the pH of the mixture.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Remember, pH is defined as the negative log of the hydronium ion concentration.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Which is the negative log of 0.0273
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
2 significant figures
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
, which comes out to 1.56.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
2 significant figures
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
In a pH, the digits to the right of the decimal are significant. So this answer has 2
significant figures.
mol H initial
0.150 mol H 2SO4
2 mol H 


 0.125 L  0.0375 mol H 
1L
1mol H 2SO4
mol OH initial
0.200 mol KOH 1mol OH 


 0.150 L  0.0300 mol OH 
1L
1mol KOH
Excess mol H +  0.0375 mol H   0.0300 mol OH   0.0075 mol H 


0.0075
mol
H
0.0075
mol
H
 H 3O +    H +  
 0.0273 M

    0.125 L  0.150 L  
0.275 L
pH   log  H 3O     log  0.0273   1.56
pH of Final Mixture
125.0 mL of 0.150 M H2SO4 is mixed with 150.0 mL of 0.200 M KOH.
What is the pH of the final mixture?
Now we have answered the original question. The pH of the final mixture is 1.56. This low value means
the solution is fairly acidic. This is reasonable because a strong acid is in excess in this case.