Guide for the Acid-Base-Program: Actph

5d. Guide for the Acid-Base-Program: ActpH

To understand this guide, it is necessary in the first place to read the “Guide for the Acid-Base-Program: ConcpH” which contains basic rules and also is useful for this this guide.

In this program the activities will be calculated and for that reason not only the concentrations of the ions are used but also their hydrated diameters. Also ions which do not participate in hydrolysis (non-hydrolytic) must be taken into consideration when calculating the ionic strength. This means that all constituents of a solution in some way must be used in the calculations.

There are 3 sections where values can be input.
1) The upper row of the form contains the current values of pKw and the temperature. Notice that the temperature is only used for the activity-values and not for pKw. The known pKw value is entered for a given temperature.
2) The large section for input contains the pK-values and the values for the concentrations of hydrolytic ions.
3) The section at the bottom of the form is the input for values of the non-hydrolytic ions.
Regarding Strong 1-protic acids and Strong 1-protic bases it is important to note that the non-hydrolytic ions must be input in section 3. If we have 0.1M HCl, we put in 0.1 in the editor-field for HA in section 2 but also the charge and concentration of Cl- in section 3. If we have 0.1M NaOH,we put in 0.1 in the editor-field for MeOH in section 2 and the charge and concentration of Na+ in section 3.
Regarding all other pure acids and bases we can do all the inputs of their values in section 2.

Regarding salts we have to divide them as has been shown in the guide for ConcpH but also to take the non-hydrolytic ions into consideration. If we have NH4KSO4 we must divide it into NH4+ K+ and SO42- . The values of NH4+ and SO42- will be put in in section 2 and K+ in section 3.

In section 2 we must also put in the values for the ion diameters. If this is not done, there will only be a calculation of ConcpH and an “Error message” will be visible. If we put in the missing values, we can press for calculation and have the value of ActpH.

The temp (default 25oC) can be changed but this only affects Davies equation but not the pK-values which must be found in tables.

In the form, a value of the ionic strength will be shown.

If this value >0.2M input the Davies Coeff = 0.15 and if I >0.6M, the value of ActpH can be unreliable since it exceeds the limit of the Davies modification of the Debye-Huckel equation.

Pure acids and bases

Ex1. Calculate pH for 0.01M HF. This is a WeakAcid
1. WeakHA : pK1=3.25. HA = 0.01 , A- = 0 and d(A-) = 3
Answer ConcpH=2.676 ActpH=2.678

If scrolling down we 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.
NaOH is a strong base and for that reason the value for Na+ will
be put in section 3 while HF is a weak acid and HF will be input in
section 2.

1.WeakHA: HA=0.01 , A-=0 and d(A-)=3.
1.StrongBase MeOH: MeOH=0.02

Section 3: Ion=Na; Charge=+1; d(Na+)=4; Conc =0.02

Answer: ConcpH=12.000 ActpH=12.051
Ex 3. Calculate pH for a mixture of 1.52e-5M HCl and 3.61e-6M NaOH
This is a strong 1-protic strong acid and a strong 1-protic base. For
both of them we must use input in section 3.
StrongHA: HA=1.52e-5 Strong Base MeOH: MeOH=3.61e-6
Section3: Ion=Cl- ; Charge=–1; Conc=1.52e-5; Ion=Na+ ; Charge=+1; d(Na+)=4 Conc=3.61e-6
Answer ConcpH=4.936 ActpH=4.938
Ex 4 Calculate pH for a mixture of 0.02M H2SO4 , 0.03M H3PO4 and 0.04 M NH3.

H2SO4 : 1.StrongH2A: pK2=2.00 H2A=0.02 HA-=0 d(HA-)=4
A2-=0 d(A2-)=4
H3PO4 : 1.StrongH3A: pK1=2.15 pK2=7.21 pK3=12.36 H3A=0.03
H2A-= 0 d(H2A-)=4 HA2-=0 d(HA2-)=4 A3-=0 d(A3-)=4
NH3 : 1.WeakBase B: pK1=4.76 B=0.04 BH+=0 d(BH+)=3
Answer: ConcpH=2.158 ActpH=2.088

Salts with non-protolytic ions

Ex 5. Calculate pH for 0.1M NaHSO3. The salt is totally dissociated into Na+ and HSO3-.
HSO3-: 1.Weak H2A: pK1=1.89 pK2=7.20
H2A=0, HA- = 0.1, d(HA-)=4, A2- = 0, d(A2-)=4.
Section3: Ion: Na Charge +1 Conc=0.1 d(Na+)=4
Answer: ConcpH=4.571 ActpH=4.357

Ex6. Calculate pH for 0.2M Na2SO3. The salt is totally dissociated into 2Na+ and SO32-
SO32- : 1.Weak H2A: pK1=1.89 pK2=7.20
H2A=0 HA-=0 d(HA-)=4 A2-=0.2 d(A2-)=4
Section 3: Ion=Na; Charge =+1; d(Na+)=4; Conc =0.4 .

Answer: ConcpH=10.250 ActpH=9.972
Note: The Ionic strength is 0.5999 which is the limit of the Davies modification of

the Debye-Huckel equation. A Davies coeff of 0.15 was used since the ionic strength exceeded 0.2.
Ex 7 Calculate pH for a mixture of 0.01M NaH2PO4 , 0.02M K2HPO4 and 0.03M NH4Cl
The salts can be divided in the following manner
0.01M Na+ +0.01M H2PO4- 0.04M K+ + 0.02M HPO42- 0.03M NH3 + 0.03M HCl
H3PO4 : 1. Weak H3A pK1= 2.15 , pK2=7.21 , pK3=12.36

H3A = 0 , H2A- = 0.01, d(H2A-)=4 , HA2- = 0.02, d(HA2-)=4,
A3-=0, d(A3-)=4
NH4+ : 1. Weak B pK1=4.76 , B = 0 , BH+ = 0.03, d(BH+)=3
Section 3: Ion Na; Charge +1; Conc= 0.01; d(Na+)=4 Ion=K; Charge =+1; Conc=0.04; d(K+)=3
Ion=Cl; Charge=-1; Conc= 0.03 d(Cl-)=3
Answer: ConcpH = 7.478 ActpH = 7.162

c) Salts with protolytic ions

Ex 8. Calculate pH for 0.15M NH4CN
0.15M NH4CN can be divided 0.15M NH3 and 0.15M HCN
0.15M NH3: 1. Weak Base B: pK1=4.76 B = 0.15 BH+ = 0 d(BH+)=3

0.15M HCN: 1. Weak HA pK1=9.22 HA = 0.15 and A- = 0 d(A-)=3

Answer ConcpH = 9.230 ActpH=9.331
Ex 9. Calculate pH for the mixture 0.025M NaHSO4 and 0.012M HAc at 50o C
Put in pKw=13.262 (from table) temp=50 in the upper part of the form
0.025M NaHSO4 can be divided into 0.025M Na+ and 0.025M HSO4-
0.025M Na+ in section 3: Ion=Na Charge=+1 Conc=0.025; d(Na+)=4
HSO4- :1.StrongH2A: pK2=1.92 H2A=0 HA-=0.025 d(HA-)=4 A2-=0 d(A2-)=4
Note we have entered pK2 for 500C for H2SO4.

HAc: 1.Weak HA : pK1=4.76 HA=0.012 A-=0 d(A-)=4
Answer: ConcpH=1.908 ActpH=1.872

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 into 0.024M NH4+, 0.012M Li+ and 0.012M
PO43-

0.024M NH4+: input in WeakBase B ; 0.012M Li+ input in section 3
0.012M PO43-: input in 1.WeakH3A with HA3-=0.012

0.020M NaH2PO4 can be divided into 0.020M Na+ and 0.020M H2PO4-
0.020M Na+ input in section 3

0.020M H2PO4-: input in 1.WeakH3A with H2A-=0.020

0.013M K2HAsO4 can be divided into 0.026M K+ and 0.013M HAsO42-
0.026M K+ input in section3
0.013M HAsO42-: input in 2.WeakH3A with HA2-=0.013
0.0021M NaOH: input in StrongBase B
0.0021M Na+ input in section 3
1.Weak H3A: pK1=2.15 pK2=7.21 pK3=12.36 H3A=0 H2A-=0.020
d(H2A-)=4
HA2-=0 d(HA2-)=4 A3-=0.012 d(A3-)=4
2.Weak H3A pK1=2.25 pK2=7.00 pK3=11.52 H3A=0 H2A-=0 d(H2A-)=4
HA2-=0.013 d(HA2-)=4 A3-=0 d(A3-)=4

3.Weak Base B pK1 = 4.76 B=0 BH+=0.024 d(BH+)=3

StrongBaseB: MeOH=0.0021
Section 3: Ion=Li Charge=+1 d(Li)=6 Conc=0.012 Ion=Na Charge=+1 d(Na+)=4 Conc=0.020 +0.0021=0.0221
Ion= K Charge=+1 Conc=0.026 d(K+)=3
Answer: ConcpH=7.895 ActpH=7.591

Calculation of titration

Ex 11. 10.00 ml 0.015M H3PO4 is to be titrated with 0.010M NaOH. Calculate pH at
a) 1st EP b) 2nd EP c) in the midpoint between 2nd and 3rd EP.
H3PO4: pK1=2.15 pK2=7.21 pK3=12.36 all d-values = 4
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, d(H2A-)=4, HA2-=0, d(HA2-)=4, A3-= 0 d(A3-)=4

MeOH = 0.0060
Section 3: Ion=Na Charge=+1 Conc=0.0060 d=4
Answer: ConcpH= 4.850 ActpH=4.797
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, d(H2A-)=4, HA2-=0, d(HA2-)=4, A3-= 0 d(A3-)=4

MeOH = 0.0075

Section 3: Ion=Na Charge=+1 Conc=0.0075
Answer: ConcpH=9.358 ActpH=9.282
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, d(H2A-)=4, HA2-=0, d(HA2-)=4, A3-= 0 d(A3-)=4

MeOH = 0.00789

Section 3: Ion= Na Charge=+1 Conc=0.00789
Answer: ConcpH=11.143 ActpH= 11.145

Calculation of NIST Buffer Solutions

Ex 12.
Calculation of phosphate buffers at 0, 25 and 50 degree Celsius

0.025M KH2PO4 + 0.025M Na2HPO4 All ions but K+ have d=4 d(K+) = 3

[Na+] = 0.050 [K+] = 0.025

Temp pKw pK1 pK2 pK3 paH NIST Diff

0 14.94 2.05 7.31 12.36 6.982 6.984 -0.002

25 14.00 2.15 7.20 12.36 6.858 6.865 -0.007

50 13.26 2.26 7.18 12.36 6.825 6.833 -0.008

NOTE: The NIST buffer concentrations are given in molality, and we use molarity, but the differences will be small.

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 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.