The [H+] of a resulting solution that is 0.01 M acetic acid (Ka = 1.8 x 10-5) and 0.01 M in benzoic acid (Ka = 6.3 x 10-5) is

Question

The [H+] of a resulting solution that is 0.01 M acetic acid (Ka = 1.8 x 10-5) and 0.01 M in benzoic acid (Ka = 6.3 x 10-5) is (A) 9 x 10-4 (B) 81 x 10-4 (C) 9 x 10-5 (D) 2.8 x 10-3

Tardigrade Admin · Accepted Answer

[ textH+] = √ textk1 textC1 + textk2 textC2 (for mixture of two weak acids) = √ text1.8 × 1 0- 5 × text0.01 + text6.3 × 1 0- 5 × text0.01 = √ text18 × 1 0- 8 + 6 3 × 1 0- 8 = 9 × 1 0- 4

Q. The [H⁺] of a resulting solution that is 0.01 M acetic acid (K_a = 1.8 x 10^-5) and 0.01 M in benzoic acid (K_a = 6.3 x 10^-5) is

9 x 10^-4

81 x 10^-4

9 x 10^-5

2.8 x 10^-3

$[H^{+}] = k_{1} C_{1} + k_{2} C_{2}$
(for mixture of two weak acids)
$= 1.8 \times 1 0^{- 5} \times 0.01 + 6.3 \times 1 0^{- 5} \times 0.01$
$= 18 \times 1 0^{- 8} + 63 \times 1 0^{- 8}$
$= 9 \times 1 0^{- 4}$

Q. The [H+] of a resulting solution that is 0.01 M acetic acid (Ka = 1.8 x 10-5) and 0.01 M in benzoic acid (Ka = 6.3 x 10-5) is

9 x 10-4

81 x 10-4

9 x 10-5

2.8 x 10-3