Solveeit Logo
Login

Question

Question: The ionic product of water is \({10^{ - 14}}\) . What is the \({{\text{H}}^{\text{ + }}}\) ion conce...

The ionic product of water is 10−14{10^{ - 14}} . What is the H + {{\text{H}}^{\text{ + }}} ion concentration of a 0.010.01 M NaOH{\text{NaOH}} solution?
A.10−14{10^{ - 14}} M
B.14 M
C.13 M
D.10−12{10^{ - 12}} M

Explanation

Solution

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;
[H + ] = Kw[OH - ]\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 10−14{\text{1}}{{\text{0}}^{ - 14}} .
We need to find out the hydrogen ion concentration of a 0.010.01 M NaOH{\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, 0.010.01 M sodium hydroxide solution will indicate that 0.010.01 M is also the concentration of the hydroxyl ions. Hence, [OH - ]=0.01\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right] = 0.01 .
Now, the ionic product of water is given by Kw = [H + ][OH - ]{{\text{K}}_{\text{w}}}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right] .
So, [H + ] = Kw[OH - ]\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:
[H + ] = 10−140.01 ⇒[H + ] = 10−1410−2 ⇒[H + ] = 10−12  \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 0.010.01 M NaOH{\text{NaOH}} solution is 10−12{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:
[OH - ] = Kw[H + ]\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.