Question

Calculate the pH change that results when 12 mL of 2.8 M HCl is added to 580. mL of each of the following solutions. Use the Acid-Base Table. Please see pH changes - part a and b and pH changes - part c and d for assistance.

(a) pure water

(b) 0.10 M CH3COO−

(c) 0.10 M CH3COOH

(d) a solution that is 0.10 M in each CH3COO− and CH3COOH.

Answer 1

amount of acid added:

V = 12 mL; [HCl] = 2.8 M

V = 580 mL of solution...

(a) pure water

mmol of HCl = MV = 2.8*12 = 33.6 mmol of H+

total V

V = 580+12 = 592 mL

[HCl] = mmol/V = 33.6 /592 = 0.056756

[H+] = 0.056756

pH = -log(0.056756) = 1.245

(b) 0.10 M CH3COO−

CH3CO- + H+

mmol of CHCOO - = 0.1*580 = 58

after adding mmol of H+ = 33.6

there is reaction

CHCOO- = 58-33.6 = 24.4

CHCOOH = 33.6

this is a buffer

pH = pKa + log(A-/HA)

pKa = 4.75

pH = 4.75 + log(24.4/33.6)

pH = 4.611

(c) 0.10 M CH3COOH

CH3COOH <--> CH3COO- + H+

Ka = [CH3COO-][H+]/[CH3COOH]

Vtotal = 580+12 = 592 mL

[CH3COO-] = 0

[H+] = 0.056756

[CH3COOH] = 0.1*580/592 = 0.097972

in equilbrium

[CH3COO-] = 0 + x

[H+] = 0.056756 + x

[CH3COOH] = 0.1*580/592 = 0.097972- x

1.8*10^-5 = x*(0.056756 +x) /(0.097972-x)

(1.8*10^-5)(0.097972) = 0.056756 x+ x^2

x^2 + 0.056756 x - 0.00000176349 = 0

x = 3.1*10^-5

[H+] = 0.056756 + x

[H+] = 0.056756 + 3.1*10^-5

ph = -log(0.056787) = 1.2457

(d) a solution that is 0.10 M in each CH3COO− and CH3COOH

mmol of CH3COO- = 0.1*580= 58

mmol of CH3COOH = 0.1*580 = 58 mmol

mmol of HCl = MV = 2.8*12 = 33.6 mmol of H+

mmol of CH3COO- = 58-33.6 =24.4

mmol of CH3COOH = 58 +33.6 = 91.6

pH = pKa + log(CH3COO-/CH3COOH)

pH = 4.75 + log(24.4 /91.6)

ph = 4.1754