Question

What is the pH of \[1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}}\] solution of \[NaOH\]?

Answer 1

Hint: The pH of a strong base is always greater than \[7\] and \[NaOH\] is a strong base. The strong base or strong acid dissociates completely into its ions in the ratio of \[1:1\]. Therefore with the help of an ionic product of water we can find the concentration of \[{H^ + }\] ions and thus we will find the pH of the given base.
Formula Used:
\[\left[ {{H^ + }} \right]{\text{ }}\left[ {O{H^{ - 1}}} \right]{\text{ = 1}}{{\text{0}}^{ - 14}}\]

Complete answer:
According to pH scale the compounds having pH value greater than seven are considered to be base and compounds having pH value less than seven are known as acids. The pH of a strong base is always greater than seven. \[NaOH\] is also a strong base therefore its pH will be greater than seven. When \[NaOH\] gets dissociated it dissociates into \[1:1\] molar ratio of its ions. The dissociation of \[NaOH\] can be shown as:
\[NaOH{\text{ }} \to {\text{ N}}{{\text{a}}^ + }{\text{ + O}}{{\text{H}}^ - }\]
Since the molar concentration of \[NaOH\] is \[1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}}\] , therefore we can write as:
\[\left[ {NaOH} \right]{\text{ = }}\left[ {{\text{N}}{{\text{a}}^ + }} \right]{\text{ = }}\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }}1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}}\]
Hence the concentration of hydroxide ions is \[1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}}\].
Now we will use the ionic product of water and then find the concentration of hydronium ions. According to ionic product of water we now that,
\[ \Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ }}\left[ {O{H^{ - 1}}} \right]{\text{ = 1}}{{\text{0}}^{ - 14}}\]
\[ \Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ = }}\dfrac{{{\text{1}}{{\text{0}}^{ - 14}}}}{{\left[ {O{H^{ - 1}}} \right]}}\]
On substituting the values we get the result as:
\[ \Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ = }}\dfrac{{{\text{1}}{{\text{0}}^{ - 14}}}}{{1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ }}}}\]
\[ \Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ = 7}}{\text{.14 }} \times {\text{ 1}}{{\text{0}}^{ - 13}}{\text{ M}}\]
We know that, \[pH{\text{ = - log}}\left[ {{H^ + }} \right]\] , therefore we can write as:
\[ \Rightarrow {\text{ }}pH{\text{ = - log}}\left[ {{\text{ 7}}{\text{.14 }} \times {\text{ 1}}{{\text{0}}^{ - 13}}} \right]\]
Using the property: \[\log \left[ {a \times b} \right]{\text{ = log}}\left[ a \right]{\text{ + log}}\left[ b \right]\] we can write as:
 \[ \Rightarrow {\text{ }}pH{\text{ = - }}\left( {{\text{log}}\left[ {{\text{ 7}}{\text{.14}}} \right]{\text{ + log}}\left[ {{{10}^{ - 13}}} \right]} \right)\]
\[ \Rightarrow {\text{ }}pH{\text{ = - }}\left( {{\text{log}}\left[ {{\text{ 7}}{\text{.14}}} \right]{\text{ - 13log10}}} \right)\]
On solving the above equation by finding logarithmic values we get the result as:
\[ \Rightarrow {\text{ }}pH{\text{ = 13 - 0}}{\text{.85}}\]
\[ \Rightarrow {\text{ }}pH{\text{ = 12}}{\text{.15}}\]

Note:
The strong electrolyte dissociates completely into its ions, therefore \[NaOH\] dissociates completely into its ions respectively. We can also find the pH of the base by finding the pOH of base and then subtracting it from \[14\] as we know that \[{\text{pH + pOH = }}14\].