Strong Acid–Strong
Base Mixture
Calculations
Example 1
Here we’ll go over an example where a strong acid is mixed with a strong base, and we
calculate the pH of the final mixture.
150.0 mL of 0.200 M Sr(OH)2 is added
to 350.0 mL of 0.100 M HNO3.
We’re given that 150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
150.0 mL of 0.200 M Sr(OH)2 is added
to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
And we’re asked to calculate the pH of the final mixture.
Sr  OH  2  Sr 2  2OH 
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
We’ll point out something important here. Strontium hydroxide is a strong base,
Sr  OH  2  Sr 2  2OH 
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
and its formula, Sr(OH)2, has 2 OH’s in it.
Sr  OH  2  Sr 2  2OH 
Dissociation
Equation
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
So when we write the balanced dissociation equation for Sr(OH)2.
2 mol OH 

1 mol Sr  OH  2
Sr  OH  2  Sr 2  2OH 
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
We see that there are 2 moles of OH- for each mole of Sr(OH)2.
2 mol OH 

1 mol Sr  OH  2
Sr  OH  2  Sr 2  2OH 
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
So we can use this as a conversion factor
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
We’ll start by calculating the initial moles of OH minus added.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
We take 0.200 moles of Sr(OH)2 per L
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Multiply it by the conversion factor 2 mol OH- over 1 mole Sr(OH)2.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
And by 0.150 L. We rounded this to 3 significant figures to save room here. The concentration 0.200
mol/L is only 3 significant figures so the answer to this calculation is limited to 3 significant figures.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
When we multiply all three numbers we get 0.0600 mol. So the initial moles of OH minus added is
0.0600 mol. Notice, this is expressed to 3 significant figures, which is consistent with the given data.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Our next step is to calculate the initial moles of H+ added.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
The H+ comes from the strong acid nitric acid, or HNO3. Each HNO3 releases 1 proton,
so we take 0.100 mol HNO3 per L
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Times 1 mol of H+ per 1 mol of HNO3
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Times 0.350 L. Again we rounded this to 3 significant figures to save room. The number
of significant figures in the answer is limited by the 3 significant figures in 0.100 M.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
The answer comes out to 0.0350 mol. So the initial moles of H+ added is 0.0350 moles.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Notice that in preserving 3 significant figures, both of these have 4 decimal places
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial 

Excess
0.100 mol HNO 3
1mol H

 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol H initial  0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Comparing the initial moles of OH minus with the initial moles of H+, we see that we
have more moles of OH minus than of H+, so the OH minus is in excess in this case.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
The excess moles of OH minus
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Is 0.0600 mol OH minus
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
minus 0.0350 mol H+
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Which equals 0.0250 mol OH minus
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250
mol
0.0250
mol
4 decimal 
4 decimal
4 decimal
 OH   
 0.0500 M

  0.150 L  0.350
0.500
L
placesL 
places
places
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Notice the numbers we’re subtracting both have 4 decimal places, so our answer must
also have 4 decimal places.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   
3 significant

  0.150 L  0.350 L   0.500 L  0.0500 M figures
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
When expressed to 4 decimal places, this number has 3 significant figures. The zero’s to
the left of the 2 are not significant, but the zero after the 5 is.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Since the hydroxide ion is in excess, we calculate its concentration in the final mixture. It
is equal to
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
0.0250 moles
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Divided by the total volume of the solution, which is 0.150 L of Strontium hydroxide
solution,
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Plus 0.350 L of HNO3 solution.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
So the concentration of OH minus is 0.0250 moles over 0.500 L
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Which Equals 0.0500 molar
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Because we have base in excess, we can calculate the pOH
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Which is the negative log of the hydroxide ion concentration
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Or the negative log of 0.0500
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Which comes out to 1.301 . None of our data or calculations in this problem have less
than 3 significant figures, so the pOH has 3 significant figures or three decimal places.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Now we can calculate the pH
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Which is 14 minus the pOH
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Or 14 minus 1.301
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
Which is 12.699. This must has 3 significant figures so its expressed to 3 decimal places.
mol OH initial
0.200 mol Sr(OH) 2
2 mol OH 


 0.150 L  0.0600 mol OH 
1L
1mol Sr(OH) 2
mol H initial
0.100 mol HNO 3
1mol H 


 0.350 L  0.0350 mol H 
1L
1mol HNO 3
Excess mol OH   0.0600 mol OH   0.0350 mol H   0.0250 mol OH 
0.0250 mol
0.0250 mol
 OH   

  0.150 L  0.350 L   0.500 L  0.0500 M
pOH   log  OH     log  0.0500   1.301
pH  14.000  pOH  14.000  1.301  12.699 pH of final mixture
150.0 mL of 0.200 M Sr(OH)2 is added to 350.0 mL of 0.100 M HNO3.
Calculate the pH of the final mixture.
We have now answered the question we set out to answer. The pH of the final mixture is
12.699. This is relatively highly basic which is consistent with the fact that OH- is in excess.