Question

The pH of a 0.23 M solution of acid HA is found to be 3.87. What is the Ka of the acid? The equation described by the Ka value is HA(aq) + H2O(l) <=> A-(aq) + H3O+ (aq) • Report your answer with two significant figures. Provide your answer below: K=0

Answer 1

Answer-

Given,

pH of solution = 3.87

molarity of solution = 0.23 M

K_a = ?

HA (aq) + H₂O (l)  $\Leftrightarrow$ A^- (aq) + H₃O⁺

We know that,

pH = - log ( [H₃O⁺] )

So, [H₃O⁺] = 10^-pH

Put the value of pH,

[H₃O⁺] = 10^-3.87 M

[H₃O⁺] = 1.35 * 10^-4 M

HA (aq) + H₂O (l)  $\Leftrightarrow$ A^- (aq) + H₃O⁺

Using Stiochiomerty,

It can be analyzed that, for 1 mole of HA, 1 mole of A^- and H2O⁺ are produced. i.e A^- and H2O⁺ have same concentration.

So, [A^-] = [H₃O⁺] = 1.35 * 10^-4 M

Now,

For A  $\Leftrightarrow$ B + C

Acid Dissociation Constant Ka = [B][C]/[A]

So, for our equation,

Ka = [A^-]$\cdot$[H₃O⁺]/[HA]

Put the values,

Ka = [1.35 * 10^-4]$\cdot$[1.35 * 10^-4]/[0.23]

Ka = 7.92 * 10^-8 [Answer]