Question

in the laboratory, a general chemistry student measured the ph of a 0.501 m aqueous solution of formic acid, hcooh to be 2.009. use the information she obtained to determine the ( k_a ) for this acid. ( k_a(\text{experiment}) = ) submit answer retry entire group 9 more group attempts remaining

Explanation:

Step1: Calculate $[H^+]$ from pH

$[H^+] = 10^{-pH} = 10^{-2.009}$
$[H^+] \approx 9.817 \times 10^{-3}\ M$

Step2: Define dissociation equilibrium

For $\text{HCOOH}
ightleftharpoons \text{H}^+ + \text{HCOO}^-$, at equilibrium:
$[\text{H}^+] = [\text{HCOO}^-] = 9.817 \times 10^{-3}\ M$
$[\text{HCOOH}] = 0.501 - 9.817 \times 10^{-3} \approx 0.4912\ M$

Step3: Solve for $K_a$

$K_a = \frac{[\text{H}^+][\text{HCOO}^-]}{[\text{HCOOH}]}$
$K_a = \frac{(9.817 \times 10^{-3})^2}{0.4912}$

Answer:

$1.97 \times 10^{-4}$