Question

Calculate the pH of a 1.0 × 10^-7 M solution of NaOH in water.

Short answer

pH of the 1.0 × 10^-7 M NaOH= 7.2

Step-by-step solution

Deriving Formula

As the solution is dilute so amount of OH^- ions which comes from water cannot be ignored.

Positive charge = Negative Charge

In the given case,

$\left[O{H}^{-}\right]=\left[H^{+}\right]+\left[{\text{Na}}^{+}\right]$

${\text{[Na}}^{+}\text{]}=1\times {10}^{-7}$

${\text{K}}_w={\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\Rightarrow {\text{[H}}^{\text{+}}\text{]}=\frac{{\text{K}}_{\text{w}}}{{\text{[OH}}^{\text{-}}\text{]}}$

${\text{[OH}}^{-}{\text{]=[Na}}^{+}\text{]+}\frac{{\text{K}}_{\text{w}}}{{\text{[OH}}^{\text{-}}\text{]}}$

Calculation of pH for Given Solution

$\begin{array}{l}{\text{[OH}}^{-}\text{]=(1}\times {10}^{-7})\text{+}\frac{1\times {10}^{-14}}{{\text{[OH}}^{-}\text{]}}\\ {\text{[OH}}^{-}\text{]=}\frac{1\times {10}^{-7}{\text{[OH}}^{-}\text{]}+1\times {10}^{-14}}{{\text{[OH}}^{-}\text{]}}\\ \text{Invalid <msup> element}-1\times {10}^{-7}{\text{[OH}}^{-}\text{]}-1\times {10}^{-14}=0\end{array}$

Using Formula

$\begin{array}{l}\text{x}=\frac{-\text{b}\pm \sqrt{{\text{b}}^2-4ac}}{2\text{a}}\\ {\text{[OH}}^{\text{-}}\text{]=}\frac{{\text{-(-1×10}}^{\text{-7}}\text{)±}\sqrt{{{\text{(-1×10}}^{\text{-7}}\text{)}}^{\text{2}}{\text{-4(1)(-1×10}}^{\text{-14}}\text{)}}}{\text{2(1)}}\\ {\text{[OH}}^{-}\text{]}=1.62\times {10}^{-7}\end{array}$

${\text{[H}}^{\text{+}}\text{]}=\frac{{\text{K}}_{\text{w}}}{{\text{[OH}}^{\text{-}}\text{]}}=\frac{1\times {10}^{-14}}{1.62\times {10}^{-7}}=6.2\times {10}^{-8}$

${\text{pH=-log[H}}^{\text{+}}]=-\log (6.2\times {10}^{-8})=7.2$