In 30 cm3 solution of (CH3COOH) 0.15M are added solution of 60cm3 of a 0.075M NaOH.Calculate pH? Ka=1.77e-4

Question

I'm having trouble with this exercise .I thought that's buffer solution, but doesn't work.

I try to switch in hydrolysis of salts but I've no idea what to do.

I need your help!

Answer 1

We have to account for the change in molarity when we mix two solutions or we have to use moles with a total volume of 90 cm³ or .09 liters.

K_A = [H⁺][Ac^-]/([HAc]

Either divide by K_w = [H⁺][OH^-] or write the expression for K_B and connect to K_A:

K_A/K_W = [Ac^-]/([HAc][OH^-])

Calculating moles of HAc and NaOH we get M_iV_i = .0045 moles for each of them

Doing an ICE table in moles, we get the following for moles at equilibrium (let x be moles of HAc reacted towards equilibrium from initial conditions)

n_HAc = .0045-x n_Ac = x n_OH = .0045-x

plugging into the Equilibrium expression we get

1.77 x 10^-4/10^-14 =( x/(.0045 - x)² ) * .09 (where we divide moles by .o9 liters to get molarity)

This is a quadratic in x that can be solved. I get x = .00499849 moles

The pH = 14 + log((.0045-x)/.09) (From pH = 14 - pOH)