Answer to Question #112375 in Inorganic Chemistry for Sheri

Question

How many grams of sodium hydroxide (MM = 40) is dissolved in 100 cm3 of 0.25 M acetic acid if the pH is 5.0?

Answer 1

1) If we consider that Ka(CH₃COOH) = 1.75*10^-5, then

$K_a = {\frac {[CH_3COO^-][H^+]}{[CH_3COOH]}}=1.75*10^{-5}$

The initial quantity of acetic acid was

100/1000*0.25 = 0.025 moles.

Quantity of [H⁺] is

[H⁺] = 10^-pH = 10^-5 = 0.00001

Considering the amount of dissociated acid as x, we rewrite:

$K_a = {\frac {[CH_3COO^-][H^+]}{[CH_3COOH]}} = {\frac {x*0.00001}{0.025-x}} =1.75*10^{-5}$

0.00001x = 4.4*10^-7 - 1.75*10^-5x

2.75*10^-5x = 4.4*10^-7

x = 0.016 - amount of CH₃COO^- in solution. Concentration of H⁺ must be the same, because dissociation occurs according to equation:

CH₃COOH <-> CH₃COO^- + H⁺

But [H⁺] is less than this amount. The rest amount of [H⁺] formed was neutralized by NaOH. So the difference between [CH₃COO^-] and [H⁺] will be amount of NaOH added.

n(NaOH) = 0.016-0.00001 = 0.01599

m(NaOH) = 0.6396 grams.

2)(Not very correct solution) Dissociation constant of CH₃COOH was not given in the task. So, i guess, this task may be not about all that stuff written above, that may be too complex. So if it doesn't imply using of dissociation constants, just use this solution.

The initial quantity of acetic acid was

100/1000*0.25 = 0.025 moles.

Considering that acetic acid completely dissociates, [H⁺] must be 0.025 too. Now, due to adding of base, pH goes down.

pH=5, [H⁺] = 10^-pH = 10-5 = 0.00001.

The difference between initial H⁺ and [H⁺] will be amount of NaOH added.

n(NaOH) = 0.025-0.00001 = 0.02499

m(NaOH) = 0.9996 grams.