Question

In a solution containing 0.01 M HCl and 0.1 M ${H_2}C{O_3}$, ratio of $\left[ {{H^ + }} \right]$ produced from strong acid and weak acid respectively is 1. Report your answer as $\dfrac{x}{{500}}$. Given: ${K_{a1}}$ & ${K_{a2}}$ of ${H_2}C{O_3}$ are $4 \times {10^{ - 7}}$ &
$4 \times {10^{ - 11}}$

Answer 1

Since HCl is a strong acid and ${H_2}C{O_3}$ is a weak acid, the ratio of $\left[ {{H^ + }} \right]$ produced from strong acid and weak acid respectively is given in the question to be 1. ${K_a}$ is the acid ionization constant which is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. The numerical value of ${K_a}$ is used to predict the extent of acid dissociation.

Complete step by step Answer:
HCl is a strong Acid
$\left[ {{H^ + }} \right] = 0.01M$
${H_2}C{O_3}$ is a weak acid
$\left[ {{H^ + }} \right] = 2 \times {10^{ - 4}}M$
$0.01 > > 2 \times {10^{ - 4}}$
${K_{a1}}$ & ${K_{a2}}$ of ${H_2}C{O_3}$ are given as,
${K_{a1}} = 4 \times {10^{ - 7}}$
${K_{a2}} = 4 \times {10^{ - 11}}$
For HCl
$\left[ {{H^ + }} \right] = \sqrt {{K_{a1}}} .c = \sqrt 4 \times {10^{ - 7}} \times 0.1 = \sqrt 4 \times {10^{ - 8}} = 2 \times {10^{ - 4}}$
Therefore, we neglect $2 \times {10^{ - 4}}$
\left\\{ {{H^ + }} \right\\}ratio\dfrac{{0.01}}{{2 \times {{10}^{ - 4}}}} = \dfrac{1}{2} \times {10^{ - 2}} \times {10^4} = 0.5 \times {10^2}

Note: According to the Arrhenius theory of acids and bases, when an acid is added to water, it donates an ${H^ + }$ ion to water to form ${H_3}{O^ + }$ (often represented by ${H^ + }$ ). The higher the concentration of ${H_3}{O^ + }$ (or ${H^ + }$ ) in a solution, the more acidic the solution is. An Arrhenius base is a substance that generates hydroxide ions, $O{H^ - }$, in water. The higher the concentration of $O{H^ - }$ in a solution, the more basic the solution is. Also, a large ${K_a}$ value indicates a stronger acid (more of the acid dissociates) and small ${K_a}$ value indicates a weaker acid.