Question

1. (B)

Calculate the change in pH that occurs when 1.00 g of NaOH is added to 250 mL of each of the following solutions: (i) 0.200 M CH₃COOH + 0.200 M CH₃COONa; K_a= 1.75 x 10^-5.

(ii) 0.250 M HCl (iii) 0.1 M NaOH  

Answer 1

mol of NAOH = mass/MW = 1/39.997 = 0.02500 mol of NaOH

V = 250 mL = 0.25 L

M = mol/V =0.02500/0.25 = 0.1 M of NaOH

i)

to a buffer of acetic acid + acetate

pH = pKa + log(acetate/acetic acid)

pKA = -log(Ka) = 1.75*10^-5= 4.75

initially

acetic acid = 0.2

acetate = 0.2

after 0.1 of NaOH:

acetic acid = 0.2-0.1 = 0.1

acetate = 0.2+0.1 = 0.3

pH = 4.75 + log(0.3/0.1)

pH = 5.22712

ii)

if we add it t .25 M of HCl

so:

0.25 M of H+ + 0.10 M OH-

[H+] = 0.25-0.10= 0.15 Mo f H+

[H+] = -log(0.15) = 0.8239

iii)

total NaOH = 0.1 +0.1 = 0.20

pOH = -log(0.20)

pH = 14-0.69897

pH = 13.30103