Question

Which of the following solutions has the lowest pH? 0.002-M HClO4, 0.20-M CH3COOH, 0.02-M HCl, 2.0-M NaCl, 0.200-M HCOOH (I asked this before but I am unsure if the answer I received is correct so I am resubmitting it.)

Answer 1

a) 0.002 M HClO₄ is a strong acid and ionizes completely.

HClO₄ (aq) --------> H⁺ (aq) + ClO₄^- (aq)

[H⁺] = [HClO₄] = 0.002 M.

pH = -log (0.002)

= 2.699 ≈ 2.7

b) CH₃COOH is a weak acid and ionizes partially.

CH₃COOH (aq) <=====> H⁺ (aq) + CH₃COO^- (aq)       K_a = 1.8*10^-5

K_a = [H⁺][CH₃COO^-]/[CH₃COOH]

=====> 1.8*10^-5 = (x)(x)/(0.20 – x)

where x M is the equilibrium concentration of H⁺.

Since K_a is small, assume x << 0.20 M and write,

1.8*10^-5 = x²/0.20

=====> x² = 1.8*10^-5*0.20

=====> x² = 3.6*10^-6

=====> x = 1.8974*10^-3

[H⁺] = 1.8974*10^-3 M

pH = -log [H⁺]

= -log (1.8974*10^-3 M)

= 2.7218 ≈ 2.7

c) 0.02 M HCl is a strong acid and ionizes completely.

HCl (aq) --------> H⁺ (aq) + Cl^- (aq)

[H⁺] = [HCl] = 0.02 M.

pH = -log (0.02)

= 1.699 ≈ 1.7

d) NaCl is a neutral salt and hence, the pH = 7.0.

e) HCOOH is a weak acid and ionizes partially.

HCOOH (aq) <=====> H⁺ (aq) + HCOO^- (aq)                K_a = 1.8*10^-4

K_a = [H⁺][HCOO^-]/[HCOOH]

=====> 1.8*10^-4 = (x)(x)/(0.20 – x)

where x M is the equilibrium concentration of H⁺.

Since K_a is small, assume x << 0.20 M and write,

1.8*10^-5 = x²/0.20

=====> x² = 1.8*10^-4*0.20

=====> x² = 3.6*10^-5

=====> x = 6.0*10^-3

[H⁺] = 6.0*10^-3 M

pH = -log [H⁺]

= -log (6.0*10^-3 M)

= 2.2218 ≈ 2.2

Therefore, 0.02 M HCl will have the lowest pH (ans).