[SpartaIam]

A buffer solution of 100 mL is 0.199 M Na₂HPO₄(aq) and 0.138 M KH₂PO₂(aq). What is the change in pH resulting from the addition of 69.1 mL of 0.01 M NaOH(aq) to this buffer solution?

I tried it five times already and keep getting the same answer of -0.05 change in pH. I'm using 6.2 x 10⁸ as the ionization constant in my K_a formula. What am I doing wrong?

[Borek]

6.2x10⁸ or 6.2x10^-8?

Show your work. It is hard to say what you did wrong not knowing what you did.

[SpartaIam]

6.2x10^-8

Okay, I first found out the initial pH by setting up the reaction as

H₂PO₄^- + H₂O --><-- H₃O⁺ + HPO₄^2-

And by using the K_a value of 6.2x10^-8. I figured out that K_a equals [H₃O⁺][HPO₄^2-]/[H₂PO₄^-].

which would be 6.23x10^-8 = [H₃O⁺] (0.199)/ (0.138) which would equal 4.32x10^-8. By using pH =-log[H₃O⁺], this would figure out to 7.37 as the initial pH.


Then I figured out the 0.01 M x 0.0691 L = 6.91x10^-4 mol NaOH which would figure out to 6.91x10^-4 mol of OH- . I then figured out that the OH- would be the limiting reagent by figuring out the amount of mols in phosphoric acid (0.0138 mol). then I set up my reaction table to be

OH- + H2PO4 --><-- HPO4 + H2O
6.91x10^-4   0.0138 mol     0.0199 mol
-6.91x10^-4  -6.91x10^-4   +6.91x10^-4

0                  0.0131 mol         0.021 mol


then I reestablished equilibrium by setting up the reaction for Kc to equal Ka

H2PO4 + H2O --><-- H3O + HPO4
0.002 M                    0       0.004 M
-x                           +x       +x
0.002 -x                    x       0.004+x

Kc=Ka=(x)(0.004+x)/(0.002-x)

6.2x10^-8 = x(0.004+x)/(0.002-x)  [by ruling out that x is negligible, I removed it]

6.2x10^-8 = x(0.004)/(0.002)

x =  3.1x10^-8 = [H3O+]

pH = -log[H3O+] = -log(3.1x10^-8) = 7.51

deltapH = 7.51 - 7.37 = 0.14

(okay, I get this answer now, but I still don't know if it's right or not)

[Borek]

You don't need ICE table, all you need here is Henderson-Hasselbalch equation.