Question

Calculate the accurate pH of ${\text{5 }} \times 1{0^{ - 3}}{\text{ M}}$ formic acid solution. ${{\text{K}}_{\text{a}}}{\text{(HCOOH) = 2 x 1}}{{\text{0}}^{{\text{ - 4}}}}$

Answer 1

Hint: We should be aware of the fact that in this case, if HA is the given acid that will dissociate then, concentration of \[{{\text{H}}^ + }\] will be equal to concentration of anion ${{\text{A}}^ - }$.

Complete step by step answer:
We know that formula ${{\text{K}}_{\text{a}}}\;{\text{ = }}\dfrac{{{\text{[}}{{\text{H}}^ + }]{\text{.[}}{{\text{A}}^ - }{\text{]}}}}{{{\text{[HA]}}}}$
Where ${\text{[}}{{\text{H}}^ + }]$is the concentration of cation, ${\text{[}}{{\text{A}}^ - }{\text{]}}$ is the concentration of anion and ${\text{[HA]}}$ is the concentration of acid.

Formic acid will dissociate as –
$HCOOH \leftrightharpoons H^+ + COOH^-$

In dissociation of formic acid, we have, ${{\text{K}}_{\text{a}}}\;{\text{ = }}\dfrac{{{\text{[}}{{\text{H}}^ + }]{\text{.[COO}}{{\text{H}}^ - }{\text{]}}}}{{{\text{[HCOOH]}}}}$

Here, we know $[{{\text{H}}^ + }{\text{] = [}}{{\text{A}}^ - }{\text{]}}$ i.e. ${\text{[}}{{\text{H}}^ + }]{\text{ = [COO}}{{\text{H}}^ - }]$

So, we can write it as - ${{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}^{\text{2}}}$ i.e. ${{\text{K}}_{\text{a}}}{\text{ = }}\dfrac{{{{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}}^{\text{2}}}}}{{{\text{[COO}}{{\text{H}}^{\text{ - }}}{\text{]}}}}$

Rearranging this equation as - ${{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}^{\text{2}}}{\text{ = }}{{\text{K}}_{\text{a}}}{\text{ }} \times {\text{ [COO}}{{\text{H}}^ - }]$
\[{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}^{}}{\text{ = }}\sqrt {{{\text{K}}_{\text{a}}}{\text{ }} \times {\text{ [COO}}{{\text{H}}^ - }]} \]

Substituting the given values of dissociation constant and ${\text{[COO}}{{\text{H}}^ - }]$ we get,
\[{{\text{[}}{{\text{H}}^{\text{ + }}}{\text{]}}^{}}{\text{ = }}\sqrt {2{\text{ }} \times {\text{ 1}}{{\text{0}}^{^{ - 4}}} \times 5\; \times {\text{ 1}}{{\text{0}}^{ - 3}}} \]
\[{\text{[}}{{\text{H}}^{\text{ + }}}{\text{] = 1}}{{\text{0}}^{ - 4}}\]
Now we know that, ${\text{ pH = - log [}}{{\text{H}}^{\text{ + }}}{\text{]}}$
                                  \[{\text{ pH = - log [1}}{{\text{0}}^{ - 4}}{\text{]}}\]
                                   \[{\text{pH = 4}}\]

Hence, an accurate pH of ${\text{5 }} \times 1{0^{ - 3}}{\text{ M}}$ formic acid solution is 4.

Additional information:
Ionic substances dissociate into their substituent ions in polar solvents. The ions formed are always in equilibrium with its undissociated solute in the solution. The conversion of reactants to products is always less than 100% and hence we can say that the reactants and products coexist in equilibrium. The extent to which a particular substance in solution is dissociated into ions, equal to the product of the concentrations of the respective ions divided by the concentration of the undissociated molecule is called the dissociation constant of that substance. It is also considered as a measure to determine the strength of an acid in a solution.

Note: An acid is said to be ‘strong’ when the concentration of its undissociated ions is very low and cannot be measured.