Question

A weak monoprotic acid has molar mass 180 g/mol. When 1.00 g of this acid is dissolved in enough water to obtain a 300 mL solution, the pH of the resulting solution is found to be 2.62. What is the value of Ka for this acid

Answer 1

Answer: The value of $K_a$ for the given acid is $3.58\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Initial mass of weak monoprotic acid = 1.00 g

Molar mass of weak monoprotic acid = 180 g/mol

Volume of solution = 300 mL

Putting values in above equation, we get:

$\text{Molarity of weak monoprotic acid}=\frac{1.00\times 1000}{180\times 300}\\\\\text{Molarity of weak monoprotic acid}=0.0185M$

To calculate the hydrogen ion concentration for given pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

pH = 2.62

Putting values in above equation, we get:

$2.62=-\log[H^+]$

$[H^+]=10^{-2.62}=2.40\times 10^{-3}M=0.0024M$

The chemical equation for the dissociation of weak monoprotic acid (HA) follows:

                            $HA\rightleftharpoons H^++A^-$

Initial:              0.0185

At eqllm:        0.0185-x     x     x

Evaluating the value of 'x'

$\Rightarrow x=0.0024$

So, equilibrium concentration of HA = (0.0185 - 0.0024) = 0.0161 M

Equilibrium concentration of $A^-$ = x = 0.0024 M

The expression of $K_a$ for above equation follows:

$K_a=\frac{[H^+][A^-]}{[HA]}$

Putting values in above equation, we get:

$K_a=\frac{(0.0024)\times (0.0024)}{0.0161}=3.58\times 10^{-4}$

Hence, the value of $K_a$ for the given acid is $3.58\times 10^{-4}$