Question

Calculate the pH of $\dfrac{N}{{1000}}$ sodium hydroxide $\left( {{\text{NaOH}}} \right)$ solution assuming complete ionization.

Answer 1

We know that the degree of alkalinity or acidity of a solution is known as its pH. pH is the negative logarithm of the hydrogen ion concentration. We are given the concentration of sodium hydroxide $\left( {{\text{NaOH}}} \right)$ solution. From the concentration of sodium hydroxide $\left( {{\text{NaOH}}} \right)$ calculate the concentration of hydroxide ions. Then calculate the pH.

Complete solution:
We are given that the concentration of sodium hydroxide $\left( {{\text{NaOH}}} \right)$ solution is $\dfrac{N}{{1000}}$.
$\dfrac{N}{{1000}}$ means $0.001{\text{ N}}$ solution. But ${\text{NaOH}}$ is a monoacidic base. Thus, normality of the solution is equal to the molarity of the solution.
Thus, the molarity of $\dfrac{N}{{1000}}$ sodium hydroxide $\left( {{\text{NaOH}}} \right)$ solution is $0.001{\text{ M}}$.
We are given that sodium hydroxide $\left( {{\text{NaOH}}} \right)$ solution undergoes complete ionization.
The ionisation reaction of sodium hydroxide $\left( {{\text{NaOH}}} \right)$ is as follows:
${\text{NaOH}} \rightleftharpoons {\text{N}}{{\text{a}}^ + } + {\text{O}}{{\text{H}}^ - }$
From the reaction we can see that ${\text{NaOH}}$ undergoes complete ionisation. Thus, ${\text{[NaOH]}} = [{\text{O}}{{\text{H}}^ - }]$.
Calculate the pOH using the equation as follows:
${\text{pOH}} = - \log [{\text{O}}{{\text{H}}^ - }]$
Substitute $0.001{\text{ M}}$ for the concentration of ${\text{O}}{{\text{H}}^ - }$ ions. Thus,
${\text{pOH}} = - \log [0.001{\text{ M}}]$
${\text{pOH}} = 3$
Thus, the pOH of $\dfrac{N}{{1000}}$ sodium hydroxide $\left( {{\text{NaOH}}} \right)$ solution is 3.
Calculate the pH using the equation as follows:
${\text{pH}} + {\text{pOH}} = - 14$
${\text{pH}} = 14 - {\text{pOH}}$
Substitute 3 for the pH of the solution. thus,
${\text{pH}} = 14 - 3$
${\text{pH}} = 11$

Thus, the pH of $\dfrac{N}{{1000}}$ sodium hydroxide $\left( {{\text{NaOH}}} \right)$ solution is 11.

Note: If the pH of the solution is less than 7 then the solution is acidic in nature. If the pH of the solution is equal to 7 then the solution is neutral in nature. If the pH of the solution is more than 7 then the solution is basic in nature. Here, the pH is 11 thus, we can say that the solution is basic or alkaline in nature.

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