Question

Calculate the pH of 5.0 × 10^-8 M HNO₃.

Short answer

pH of the 5.0 × 10^-8 M HNO₃= 6.9

Step-by-step solution

Deriving Formula 

As the solution is dilute so amount of H⁺ ions which comes from water cannotbe ignored.

Positive charge = Negative Charge

In the given case,

$\left[H^{+}\right]=\left[O{H}^{-}\right]+\left[{{\text{NO}}_{\text{3}}}^{-}\right]$

${{\text{[NO}}_{\text{3}}}^{-}\text{]}=5\times {10}^{-8}$

$\begin{array}{l}{\text{K}}_{\text{w}}={\text{[H}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}\\ {\text{[OH}}^{\text{-}}\text{]}=\frac{{\text{K}}_{\text{w}}}{{\text{[H}}^{\text{+}}\text{]}}\end{array}$

${\text{[H}}^{\text{+}}{{\text{]=[NO}}_{\text{3}}}^{\text{-}}\text{]+}\frac{{\text{K}}_{\text{w}}}{{\text{[H}}^{\text{+}}\text{]}}$

Calculation of pH for Given Solution 

$\begin{array}{l}{\text{[H}}^{\text{+}}\text{]=(}5\times {10}^{-8})\text{+}\frac{1\times {10}^{-14}}{{\text{[H}}^{\text{+}}\text{]}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{5\times {10}^{-8}{\text{[H}}^{\text{+}}\text{]}+1\times {10}^{-14}}{{\text{[H}}^{\text{+}}\text{]}}\\ {\left[H^{-}\right]}^2-5\times {10}^{-8}{\text{[H}}^{\text{+}}\text{]}-1\times {10}^{-14}=0\end{array}$

Using Formula

$\begin{array}{l}\text{x}=\frac{-\text{b}\pm \sqrt{{\text{b}}^2-4\text{ac}}}{2\text{a}}\\ {\text{[H}}^{\text{+}}\text{]=}\frac{{\text{-(-5×10}}^{\text{-8}}\text{)±}\sqrt{{{\text{(-5×10}}^{\text{-8}}\text{)}}^{\text{2}}{\text{-4(1)(-1×10}}^{\text{-14}}\text{)}}}{\text{2(1)}}\\ {\text{[H}}^{\text{+}}\text{]}=1.28\times {10}^{-7}\end{array}$

${\text{pH=-log[H}}^{\text{+}}]=-\log (1.28\times {10}^{-7})=6.9$