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

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.