
The ionic product of water is ${10^{ - 14}}$ . What is the ${{\text{H}}^{\text{ + }}}$ ion concentration of a $0.01$ M ${\text{NaOH}}$ solution?
A.${10^{ - 14}}$ M
B.14 M
C.13 M
D.${10^{ - 12}}$ M
Answer
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Hint: The ionic product of water is the product of the concentration of the hydrogen ion and the concentration of the hydroxyl ion.
In case of an acidic or a basic solution, the value of the ionic product of water remains the same at a given temperature.
Knowing the hydroxyl ion concentration, the hydrogen ion concentration can be calculated as follows;
$\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}$
Complete step by step answer:
The ionic product of water is given to be ${\text{1}}{{\text{0}}^{ - 14}}$ .
We need to find out the hydrogen ion concentration of a $0.01$ M ${\text{NaOH}}$ solution.
Molarity of a solution describes the number of moles of the solute per litre of the solution and is thus, an expression of the concentration of the solution.
Also, sodium hydroxide is a strong electrolyte and so it will undergo dissociation completely in solution. Also, sodium hydroxide is a base and so its concentration will give the concentration of the hydroxyl ions of the sodium hydroxide solution.
Thus, $0.01$ M sodium hydroxide solution will indicate that $0.01$ M is also the concentration of the hydroxyl ions. Hence, $\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right] = 0.01$ .
Now, the ionic product of water is given by ${{\text{K}}_{\text{w}}}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]$ .
So, $\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}$
Substitute all the values. Then we will get:
$
\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{10}^{ - 14}}}}{{0.01}} \\
\Rightarrow \left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 2}}}} \\
\Rightarrow \left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}{10^{ - 12}} \\
$
So, the hydrogen ion concentration of a $0.01$ M ${\text{NaOH}}$ solution is ${10^{ - 12}}$ M .
Hence option D is correct.
Note:
Similarly, if the value of the concentration of the hydrogen ion is given, then the value of the concentration of the hydroxyl ion can be calculated as:
$\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}}$
As the degree of ionization is proportional to temperature, the value of the ionic product also increases with the increase in temperature.
In case of an acidic or a basic solution, the value of the ionic product of water remains the same at a given temperature.
Knowing the hydroxyl ion concentration, the hydrogen ion concentration can be calculated as follows;
$\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}$
Complete step by step answer:
The ionic product of water is given to be ${\text{1}}{{\text{0}}^{ - 14}}$ .
We need to find out the hydrogen ion concentration of a $0.01$ M ${\text{NaOH}}$ solution.
Molarity of a solution describes the number of moles of the solute per litre of the solution and is thus, an expression of the concentration of the solution.
Also, sodium hydroxide is a strong electrolyte and so it will undergo dissociation completely in solution. Also, sodium hydroxide is a base and so its concentration will give the concentration of the hydroxyl ions of the sodium hydroxide solution.
Thus, $0.01$ M sodium hydroxide solution will indicate that $0.01$ M is also the concentration of the hydroxyl ions. Hence, $\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right] = 0.01$ .
Now, the ionic product of water is given by ${{\text{K}}_{\text{w}}}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]$ .
So, $\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}$
Substitute all the values. Then we will get:
$
\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{10}^{ - 14}}}}{{0.01}} \\
\Rightarrow \left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 2}}}} \\
\Rightarrow \left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}{10^{ - 12}} \\
$
So, the hydrogen ion concentration of a $0.01$ M ${\text{NaOH}}$ solution is ${10^{ - 12}}$ M .
Hence option D is correct.
Note:
Similarly, if the value of the concentration of the hydrogen ion is given, then the value of the concentration of the hydroxyl ion can be calculated as:
$\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{ = }}\dfrac{{{{\text{K}}_{\text{w}}}}}{{\left[ {{{\text{H}}^{\text{ + }}}} \right]}}$
As the degree of ionization is proportional to temperature, the value of the ionic product also increases with the increase in temperature.
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