5b. Guide for the Acid-Base-Program: ConcpH
With this program it is possible to calculate pH-value for separate acids, bases and their
salts or mixtures of these substances. It is not valid if other reactions than hydrolysis occur.
To the left on the top of the form there is default pKw-value for 25oC , which can be
changed for different temperatures. Such a value can be obtained from a chemistry data book.
The acid- and base-constants are to be input in the left part and concentrations in the right
part of the form.
There is only one separate row for input of a Strong 1-protic-acid. If there is more than
one of such acids, it is nessessary to calculate the concentration of a mixture of them before
input. The same thing is true for 1-protic strong bases.
Other acids and bases can be found in two versions. If there for instance are both sulphuric
acid and selenous acid which both are Strong 2-protoic-acids, then one of them can be input in
the row for 1. Strong H2A, and the other one in the row for 2. Strong H2A.
The starting-point is that all the constants and all the concentrations have the value 0.
Acids and bases, that will be calculated on, must have all acid- and base-constants
entered in place of 0. On the other hand some of the concentrations can be 0.
Two sorts of error message can be found.
1) This program is an English version. For that reason You have to use a point(.) instead
of a decimal comma
2) If you have not put in all the necessary pK-values, this will be shown in the field:”
Error message”, down to the left on the screen.
At the starting ”Zero position” all the values will be put to 0. (pH will be put to -)
If you calculate with the same set of acids and bases and want to calculate for other
concentrations, you need not use the ’Zero position’ but only change the concentrations.
If you after a calculation want to change any one of the values of pK or if you change any
one of the participating acids and bases in the mixture, you must use the ’Zero position’ and
after that put in all the values.
The equilibrium concentrations for the substances after the reactions are shown down-side
(End Concentrations)
The following calculations may be made
The solutions can be made up of separate pure acids and bases but also of salts, which can
contain hydrolytic ions. The representation is made easier by division in three groups: a) pure
acids and bases, b) salts without hydrolytic ions and c) salts with such ions, as the solutions
technically are different.
a) Pure acids and bases
Ex1. Calculate pH for 0.01M HF .
Go to 1.Weak HA and input pK1=3.25,
HA = 0.01, A- = 0
Go to “Press for Calculating” and you get ConcpH=2.676
If scrolling down you will find HA = 7.80E-3 M and A- = 2.20E-3 M
Ex2. Calculate pH for a mixture of 0.01 M HF and 0.02 M NaOH.
In 1.WeakAcid input pK1=3.25
HA=0.01, A-=0
In StrongBase input MeOH =0.02
Answer: ConcpH=12.000
Ex 3. Calculate pH for a mixture of 1.52e-5 M HCl and 3.61e-6 M NaOH
Strong Acid HA= 1.52e-5 and Strong Base MeOH= 3.61e-6.
Answer: ConcpH=4.936
Ex4. Calculate pH for a mixture of 0.02 M H2SO4, 0.03M H3PO4 and 0.04 M NH3.
H2SO4 1. Strong H2A with pK2=2.00 , H2A=0.02, HA- =0, A2- =0
H3PO4 1. Weak H3A with pK2= 2.15, pK2=7.21, pK3=12.36
H3A=0.03, H2A- = 0, HA2- = 0, A3-=0
NH3 1. Weak Base with pK1=4.76, B=0.04, BH+ = 0
Answer: ConcpH = 2.158
Salts of acids and bases
It is important to know if hydrolytic ions are present in the acids or bases.
1) Non-hydrolytic ions can have positive charge as Na+ and K+ or negative charge as
Cl- and Br-
2) Hydrolytic ions can have positive charge as Al3+ which in water forms aqua-metalic
ions Al(H20)63+ and ammonium ions NH4+, or negative charges as CN- and S2-.
If there are only non-hydrolytic ions, you can do the inputting directly
If there are hydrolytic ions, the salts must be divided before inputting.
b) Salts with non-protolytic ions
Ex 5.
0.1M NaHSO3. The salt will totally be dissociated to Na+ and HSO3-.
HSO3- 1.WeakAcid H2A with pK1=1.89 and pK2=7.20
H2A=0, HA- = 0.1, A2- = 0. Na+ is non-hydrolytic
Answer: ConcpH= 4.571
.
Ex 6. 0.2M Na2SO3. The salt will totally be dissociated to 2Na+ and SO32-.
SO32-: 1.WeakH2A with pK1=1.89 and pK2= 7.20
H2A=0, HA- = 0, A2- = 0.2
Answer
ConcpH= 10.250
Ex 7. 0.01M NaH2PO4 , 0.02M K2HPO4 and 0.03M NH4Cl
The salts can be divided according to the following
0.01M Na+ +0.01M H2PO4- ; 0.04M K+ + 0.02M HPO42- ; 0.03M NH4+ + 0.03M
ClH3PO4 : 1.WeakH3A with pK1= 2.15 , pK2=7.21 , pK3=12.36
H3A = 0 , H2A- = 0.01, HA2- = 0.02, A3-=0
NH4+ : 1.WeakBase
pK1=4.76 , B = 0 , BH+ = 0.03
Na+ and K+ are non-hydrolytic
Answer:
ConcpH = 7.478
c) Salts with hydrolytic ions
Each of the salts with hydrolytic ions must be divided to separate parts. This can be done
in different ways.
Ex 8. 0.15M NH4CN can be divided either in a) 0.15M NH3 and 0.15M HCN
or in b) 0.15M NH4+ and 0.15M CNa) 0.15M NH3: 1.WeakBase B with pK1=4.76 B = 0.15 BH+ = 0
0.15M HCN: 1.WeakHA with pK1=9.22 HA = 0.15 A- = 0
pH=9.230
b) 0.15M NH4+ 1.WeakBase B with pK1=4.76 B=0 BH+ =0.15
0.15M CN- 1.WeakHA with pK1=9.22 HA=0 A-=0.15
pH=9.230
Answer:
ConcpH = 9.230
Ex 9. Calculate pH for a mixture of 0.025M NaHSO4 and 0.012M HAc at 50oC
(In this example, we use pK values for the acid for 250C, absent elevated temperature values)
For 500C, pKw=13.262 (from table).
0.025M NaHSO4 can be divided in 0.025M Na+ and 0.025M HSO4- ; Na+ nonhydrolytic
0.025M HSO4- will be input in 1.Strong H2A with pK2 =1.92
H2A=0 HA-= 0.025 A2-=0;
Note we have entered pK2 for 500C for H2SO4.
0.012M HAc will be input in 1.WeakHA with pK1=4.76
Aswer:
HA=0.012 A-=0
ConcpH=1.908
Ex 10. Caculate pH for the following mixture:
0.012M (NH4)2LiPO4 , 0.020M NaH2PO4 , 0.013M K2HAsO4 , 0.0021M NaOH
0.012M (NH4)2LiPO4 can be divided in 0.024M NH4+, 0.012M Li+ and 0.012M PO430.024M NH4+ : input in WeakBase B ; Li+ is non-hydrolytic
0.012M PO43- : input in 1.WeakH3A with HA3-=0.012
0.020M NaH2PO4 can be divided in 0.020M Na+ and 0.020M H2PO4Na+ is non-hydrolytic
0.020M H2PO4- : input in 1.WeakH3A with H2A-=0.020
0.013M K2HAsO4 can be divided in 0.026M K+ and 0.013M HAsO42K+ is non-hydrolytic
0.013M HAsO42- : input in 2.WeakH3A with HA2-=0.013
1.Weak H3A: pK1=2.15 pK2=7.21 pK3=12.36
H3A=0 H2A-=0.020
HA2-=0
A3-=0.012
2.Weak H3A pK1=2.25 pK2=7.00
HA2-=0.013 A3-=0
pK3=1.52
0.0021M NaOH : input in StrongBase B:
1. Weak base B: pK1=4.76 B=0
Answer:
H3A=0 H2A-=0
MeOH=0.0021
B+=0.024
ConcpH=7.895
Calculating of titration
Ex 11. 10.00ml 0.015M H3PO4 is titrated with 0.010M NaOH. Calculate pH at a)1st EP
b)2nd EP c) midway between 2nd and 3rd EP.
H3PO4 : pK1=2.15 pK2=7.21 pK3=12.36
a) 10ml 0.015M H3PO4 + 15ml 0.010M NaOH gives 0.015*10/(10+15) = 0.0060M H3PO4
and 0.010*15/(10+15)=0.0060M NaOH.
H3A=0.0060, H2A-=0, HA2-=0, A3-= 0
0.0060M NaOH, input Strong base B: MeOH=0.0060
Answer:
ConcpH= 4.850
b) 10ml 0.015M H3PO4 + 30ml 0.010M NaOH gives 0.015*10/(10+30) = 0.00375M H3PO4
and 0.010*30/(10+30)=0.0075M NaOH.
H3A=0.00375, H2A-=0, HA2-=0, A3-= 0
MeOH=0.0075
Answer:
ConcpH=9.358
c) 10ml 0.015M H3PO4 + 37.5ml 0.010M NaOH gives 0.015*10/(10+37,5) = 0.00316M
H3PO4 and 0.010*37.5/(10+37.5)=0.00789M NaOH.
H3A=0.00316, H2A-=0, HA2-=0, A3-= 0
MeOH=0.00789
Answer:
ConcpH=11.143
Add an acid or base to a solution to get a determined pH
Ex 12 In a container there are a mixture of 2.00e-5 M H2SO4 and 1.00e-3 M H2SO3. How
much must the the start concentration of NH3 in the container be to get pH=5.5 for the
new mixture?
H2SO4 : 1.Strong H2A pK2=2.00 H2A=2.00e-5 HA- =0 A2- =0
H2SO3.: 1. Weak H2A pK1= 1.89 pK2=7.20 H2A=1.00e-3 HA- =0
NH3 : 1.WeakBase B
A2- =0
pK1=4.76 B=1e-3 BH+=0
Calculate the start concentration of NH3 to give the new mixture pH = 5.5
Let the start value be
B=1e-3
BH+=0
=>
pH=4.41
increase
B=1e-2
BH+=0
=>
pH=9.81
decrease
B=1.2e-3
BH+=0
=>
pH=6.47
decrease
B=1.1e-3
BH+=0
=>
pH=6.01
decrease
B=1.05e-3
BH+=0
=>
pH=5.37
increase
B=1.06e-3
BH+=0
=>
pH=5.57
decrease
B=1.055e-3
BH+=0
=>
pH=5.47
increase
B=1.056e-3
BH+=0
=>
pH=5.49
increase
B=1.057e-3
BH+=0
=>
pH=5.51
Answer: The start concentration for NH3 = 1.06e-3 M.
How to print the result.
1. Set the Printer to "Landscape Mode".
2. Open the Form on the Screen.
3. Click "File" (left side on the top) and then "Print".
4. It can take a long time before printing.
5. After printing the upper part you have to scroll to see the lower part of the form on the
screen.
6. Click "Print".
7. After printing click "Exit".
8. Don't forget to set the normal mode on the Printer.