Question

The Ka for benzoic acid is what is 6.28 x 10-5. If 13.7 mL a solution with an analytical concentration of HNO3 of 1.65 M is added to 363.4 mL solution with an equilibrium concentration of benzoate of 0.516 M and an equilibrium concentration of benzoic acid of 0.0861M, what is the pH of the resulting solution?

Answer 1

V = 363.4 mL solution

equilibrium concentration of benzoate, [A-] = 0.516 M
equilibrium concentration of benzoic acid, [HA] = 0.0861M

Buffer Equilibria:
[HA]=[H3O+]+[A-]
Ka = [H3O+][A-]/[HA] = 6.28E-5

[HA]   = 0.0861 M
[A-]   = 0.516 M
[H30+] = Ka*[HA]/[A-] = 6.28E-5*0.0861/0.516 = 1.05E-5
pH=4.98

Change in pH:
13.7 mL of 1.65 M HNO3 will react with the basic part [A-].
[A-] + [H30+] = [HA]

Initial:
13.7 mL of 1.65 M HNO3 will decrease the amount of [A-] by 13.7*1.65 = 22.605 millimol
and increase the amount of [HA] by 22.605 millimol

Volume of solution = 363.4+13.7 = 377.1 ml

Initial concentration after adding HNO3
[HA] = (0.0861*363.4+22.605)/377.1 = 0.143 M
[A-] = (0.516*363.4-22.605)/377.1 = 0.437 M
[H3O+]=4.98

[HA] = [A-] + [H30+]

Change:
[HA]   =- x
[A-]   = + x
[H3O+] = +x

Equilibrium:
[HA]   = 0.143-x
[A-]   = 0.437 +x

Approximation as x is very small
[HA]   = 0.143-x =0.143
[A-]   = 0.437 +x = 0.437

Buffer Equilibria:

Ka = [H3O+]*[A-]/[HA]
pH = pKa + Log( [A-]/[HA] )
Ka = 6.28E-5
pKa = -Log(6.28E-5) = 4.2
pH = 4.2 + Log(0.437/0.143) = 4.68