Question

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

Short answer

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 cannot be ignored.

Positive charge = Negative Charge

In the given case,

$$\begin{gathered} \left[H^{+}\right]=\left[O{H}^{-}\right]+\left[N{O}_3^{-}\right] \\ \left[N{O}_3^{-}\right]=5\times {10}^{-8} \\ {K}_w=\left[H^{+}\right]\left[O{H}^{-}\right] \\ \left[O{H}^{-}\right]=\frac{K_w}{\left[H^{+}\right]} \\ \left[H^{+}\right]=\left[N{O}_3^{-}\right]+\frac{K_w}{\left[H^{+}\right]} \end{gathered}$$

Calculation of pH for Given Solution

$$\begin{gathered} \left[{\mathrm{H}}^{+}\right]=\left(5\times {10}^{-8}\right)\times \frac{1\times {10}^{-4}}{\left[\mathrm{H}\right]} \\ \left[{\mathrm{H}}^{+}\right]=\frac{5\times {10}^{-8}\left[{\mathrm{H}}^{+}\right]+1\times {10}^{-}}{\left[{\mathrm{H}}^{+}\right]} \\ \left[{\mathrm{H}}^{+}\right]-5\times {10}^{-8}\left[H^{+}\right] \end{gathered}$$

Using Formula

$$\begin{gathered} \mathrm{x}=\frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}} \\ \left[{\mathrm{H}}^{+}\right]=-\left(-5\times {10}^{-8}\right)\pm \frac{\sqrt{{\left(-5\times {10}^{-8}\right)}^2-4\left(1\right)\left(-1\times {10}^{-14}\right)}}{2\left(1\right)} \\ \left[{\mathrm{H}}^{+}\right]=1.28\times {10}^{-7} \\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log \left(1.28\times {10}^{-7}\right)=6.9 \end{gathered}$$