Questions & Answers

Question

Answers

Answer
Verified

We know that the acid ionization constant ${K_a}$ is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. We can use the numerical value of ${K_a}$ to determine the extent of acid dissociation. Therefore,

${K_a} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{h}}}}}$

For solving this question, first we need to formulate a dissociation equation for the salt which is \[Ca{\left( {Lac} \right)_2}\] which will be

\[Ca{(lac)_2} \rightleftharpoons {\text{ }}C{a^{2 + }}{\text{ }} + {\text{ }}2La{c^ - }\]

Now, we are given \[0.13{\text{ }}mole\]of this salt is added in \[0.50{\text{ }}litre\]solution, so

$Ca{(lac)_2} \rightleftharpoons {\text{ }}C{a^{2 + }}{\text{ }} + {\text{ }}2La{c^ - }$

\[0.13{\text{ }}X{\text{ }}2M\;\;\;\;2{\text{ }}X{\text{ }}2{\text{ }}X{\text{ }}0.13\]

From the question we can all derive a conclusion that \[La{c^ - }\]is conjugate base of a weak acid so in the presence of water it will form \[O{H^ - }\]ion, so the equation will be

$La{c^ - }{\text{ }} + {\text{ }}{H_2}0{\text{ }} \rightleftharpoons {\text{ }}HLac{\text{ }} + {\text{ }}O{H^ - }$

At equilibrium the amount of OH- ion formed can be taken as x so

At equilibrium \[0.52{\text{ }}-x\]will be Lac- and \[HLac\]and \[O{H^ - }\]concentrations will be\[x\].

We know \[{K_h}\]is equal to the product of concentration of product divided by the concentration of reactant. So,

${K_h} = \dfrac{{{{\text{x}}^{\text{2}}}}}{{{\text{0}}{\text{.52 - x}}}}{\text{ = }}\dfrac{{{{\text{x}}^2}}}{{0.52}}$ , \[x\]has been neglected in denominator because \[x < < {\text{ }}52\]

We are given the value of \[pOH\]which is 5.26 so the concentration of \[O{H^ - }\]will be

\[\left[ {O{H^ - }} \right]{\text{ }} = {\text{ }}{10^{ - 5.26}} = {\text{ }}2.56{\text{ }}X{\text{ }}{10^{ - 6}} = {\text{ }}x\]

Now, we are having the value of x, by putting it into the equation of \[{K_h}\]we get

${K_h} = \dfrac{{{{(2.56 \times {{10}^{ - 6}})}^2}}}{{0.52}} = 12.12 \times {10^{ - 12}}$

Also ${K_a} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{h}}}}}$= $\dfrac{{{{10}^{ - 14}}}}{{12.12 \times {{10}^{ - 12}}}} = \;8.26{\text{ }}X{\text{ }}{10^{ - 4}}$

We must know that the dissociation is breaking up of molecules into two ions out of which the molecule is being made. Mainly happens when the molecule is being placed in aqueous solution. This happens with salts and acids, and t is dissociation due to which the conductivity of water increases.