A buffer solution is made by adding 75.0 g soduim acetate to 500.0 mL of a 0.64 M solution of acetic acid. What is the pH of the solution?

5.2

Explanation:

CH3CO2H (aq) + H2O(l)  CH3CO2

-

(aq) + H3O+

(aq)

Initial: 0.64M 1.8M 10-7M

Ka = 1.8 x10-5 = [CH3CO2

-

][ H3O+

] / [CH3CO2H] =

Henderson-Hasselbach equation: pH=pKa + log [A-

]o/[HA]o

Concentrations of [A-

]o,[HA]o are within an order of magnitude

[A-

]o= [CH3CO2

-

] = 1.8M ; /[HA]o = [CH3CO2H] = 0.64M

pH= 4.74 + log [1.8M/0.64M] = 5.2

5.2

Explanation:

Will work problem two ways (1) applying common ion analysis and (2) using the Henderson-Hasselbalch equation.

Given 75g NaOAc into 500ml of 0.67M HOAc

=> (75g/82.034g/mol) / 0.500L = 1.829M NaOAc solution

Common-ion Analysis

             HOAc    ⇄    H⁺   +   OAc⁻

C(i)        0.67M            0         1.829M

ΔC            -x               +x           +x

C(f)         0.67-x                       1.829+x

            ≅ 0.67M          x         ≅ 1.829M

Ka = [H⁺][OAc⁻]/[HOAc] => [H⁺] = Ka[HOAc]/[OAc⁻]

=> (1.7x10⁻⁵)(0.67)/(1.829) = 6.2 x 10⁻⁶M in H⁺

pH = -log[H⁺] = - log(6.2x10⁻⁶) = 5.2

Henderson-Hasselbalch equation

pKa(HOAc) = -logKa = -log(1.7x10⁻⁵) = 4.76

pH = pKa + log[Base]/[Acid] = 1.76 + log(1.829M)/(0.67M) = 5.2