A 50.0 mL solution of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH. What is the pH after 20.0 mL of base has been added? Ka CH3COOH = 1.8 x 10-5

Question

A 50.0 mL solution of 0.150 M acetic acid (CH₃COOH) is titrated with 0.150 M NaOH.

What is the pH after 20.0 mL of base has been added? K_a CH₃COOH = 1.8 x 10^-5

Answer 1

Find the moles of both:

moles NaOH:

M=n/V

n=M(V)=(0.150)(20/1000)

n=0.003 mol NaOH

moles CH3COOH:

n=MV

n=(0.150)(50/1000)

n=0.0075 mol CH3COOH

Use the henderson equation:

pH=pKa+log([CH₃COO^-]/[CH₃COOH])

pH=pKa+log(mol CH₃COO^-/ mol CH₃COOH)

pH=pKa+log(^{mol CH3COO-+mol NaOH}/ _{mol CH3COOH-mol NaOH})

pH=-log(1.8×10^-5)+log(^0+0.003/_0.0075-0.003)

pH=4.5686