pH of resulting solution?

Question

1.08 g of NaH2PO4 and 3.66 g of K2HPO4 are dissolved in sufficient water to make a liter of solution. Calculate the pH of the resulting solution.

Answer 1

Atomic mass:
H ----1.008
O ----16.00
Na -- 22.99
P ----30.97
K ----39.10
Hence the molecular weight of NaH2PO4 is:
22.99 + 2*1.008 + 30.97 + 4*16.00 = 119.98
and the molecular weight of K2HPO4 is:
2*39.10 + 1.008 + 30.97 + 4*16.00 = 174.18
In 1 liter of solution, the initial concentration of NaH2PO4 is:
1.08/ 119.98 = 0.00900 (M)
and the initial concentration of K2HPO4 is:
3.66/ 174.18 = 0.0210 (M)
We also need following Ka values:
H3PO4(aq) <--> H+(aq) + H2PO4(-)(aq) Ka1 = 7.5 × 10^(-3)
H2PO4(-)(aq) <--> H+(aq) + HPO4(2-)(aq) Ka2 = 6.2 × 10^(-8)
HPO4(2-)(aq) <--> H+(aq) + PO4(3-)(aq) Ka3 = 1.7 × 10^(-12)
The pH is estimated approximately nutral (we will check it later), thus we do not need to involving Ka1 and Ka3. Let the final [H+] to be x M, and let us write:
---------------- H2PO4(-)(aq) <--> H+(aq) + HPO4(2-)(aq)
Initial: ------- 0.009 ------------------ 0.00 -------- 0.021
Final: ------- (0.009 - x) --------------- x -----------(0.021+x)
Hence: x(0.021 +x)/(0.009 - x) = Ka2 = 6.2 × 10^(-8)
This can be rearranged into a quadratic equation
X^2 + 0.021x - 5.58E-10 = 0
with solutions:
2.657× 10^(-8) and - 0.021
Ignore the negative one and [H+] = 2.657× 10^(-8).
Let us check to see if our approximation is valid or not. If final [H+] = 2.657× 10^(-8),
Ka1 = 7.5 × 10^(-3) = [H+] [H2PO4(-)]/ [H3PO4]
= 2.657× 10^(-8)*0.009/[H3PO4]
which leads to [H3PO4] ≈ 3.2× 10^(-8), which does not alter [H2PO4(-)] = 0.009.
Also Ka3 = 1.7 × 10^(-12) = [H+] [PO4(3-)]/ [HPO4(2-)]
= 2.657× 10^(-8)*[PO4(3-)]/ 0.021
which leads to [PO4(3-)] ≈ 1.3 × 10^(-6), which does not alter [HPO4(2-)] = 0.021.
Thus our calculation is valid and the pH of the resulting solution is:
- log[2.657× 10^(-8)] = 7.5756 ≈ 7.58