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Question: What is the pH of \[1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}}\] solution of \[N...

What is the pH of 1.4 × 10−2 M1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}} solution of NaOHNaOH?

Explanation

Solution

The pH of a strong base is always greater than 77 and NaOHNaOH is a strong base. The strong base or strong acid dissociates completely into its ions in the ratio of 1:11:1. Therefore with the help of an ionic product of water we can find the concentration of H+{H^ + } ions and thus we will find the pH of the given base.
Formula Used:
[H+] [OH−1] = 10−14\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. NaOHNaOH is also a strong base therefore its pH will be greater than seven. When NaOHNaOH gets dissociated it dissociates into 1:11:1 molar ratio of its ions. The dissociation of NaOHNaOH can be shown as:
NaOH → Na+ + OH−NaOH{\text{ }} \to {\text{ N}}{{\text{a}}^ + }{\text{ + O}}{{\text{H}}^ - }
Since the molar concentration of NaOHNaOH is 1.4 × 10−2 M1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ M}} , therefore we can write as:
[NaOH] = [Na+] = [OH−] = 1.4 × 10−2 M\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 × 10−2 M1.4{\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,
⇒ [H+] [OH−1] = 10−14\Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ }}\left[ {O{H^{ - 1}}} \right]{\text{ = 1}}{{\text{0}}^{ - 14}}
⇒ [H+] = 10−14[OH−1]\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:
⇒ [H+] = 10−141.4 × 10−2 \Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ = }}\dfrac{{{\text{1}}{{\text{0}}^{ - 14}}}}{{1.4{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 2}}{\text{ }}}}
⇒ [H+] = 7.14 × 10−13 M\Rightarrow {\text{ }}\left[ {{H^ + }} \right]{\text{ = 7}}{\text{.14 }} \times {\text{ 1}}{{\text{0}}^{ - 13}}{\text{ M}}
We know that, pH = - log[H+]pH{\text{ = - log}}\left[ {{H^ + }} \right] , therefore we can write as:
⇒ pH = - log[ 7.14 × 10−13]\Rightarrow {\text{ }}pH{\text{ = - log}}\left[ {{\text{ 7}}{\text{.14 }} \times {\text{ 1}}{{\text{0}}^{ - 13}}} \right]
Using the property: log⁡[a×b] = log[a] + log[b]\log \left[ {a \times b} \right]{\text{ = log}}\left[ a \right]{\text{ + log}}\left[ b \right] we can write as:
⇒ pH = - (log[ 7.14] + log[10−13])\Rightarrow {\text{ }}pH{\text{ = - }}\left( {{\text{log}}\left[ {{\text{ 7}}{\text{.14}}} \right]{\text{ + log}}\left[ {{{10}^{ - 13}}} \right]} \right)
⇒ pH = - (log[ 7.14] - 13log10)\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:
⇒ pH = 13 - 0.85\Rightarrow {\text{ }}pH{\text{ = 13 - 0}}{\text{.85}}
⇒ pH = 12.15\Rightarrow {\text{ }}pH{\text{ = 12}}{\text{.15}}

Note:
The strong electrolyte dissociates completely into its ions, therefore NaOHNaOH 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 1414 as we know that pH + pOH = 14{\text{pH + pOH = }}14.