Answer to Question #114058 in General Chemistry for lill

Question

The pH of a 0.6M solution of a weak acid, HA (aq), is 1.98 at 25oC. What is the value of K?
a. 6.53 b. 1.86 x 10-4 c. 1.75 x 10-3 d. 1.8 x 10-2

Answer 1

HA <-> H⁺ + A^-

If x of HA have dissociated then

K = ${\frac {[H^+][A^-]}{[HA]}}={\frac {x*x}{0.6-x}}$

[H⁺] = [A^-] = x = 10^-pH = 10^-1.98 = 0.0105

${\frac {[H^+][A^-]}{[HA]}}={\frac {x*x}{0.6-x}}={\frac {0.0105*0.0105}{0.6-0.0105}}={\frac {0.00011}{0.5895}}=1.866*10^{-4}\approx1.86*10^{-4}$

Answer b 1.86 x 10^-4