Question

Calculate the pH of 7.0 × 10^-7 M HCl solution.

Short answer

pH of the 7.0 × 10^-7 M HCl= 6.15

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,

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

${\text{[Cl}}^{-}\text{]}=7\times {10}^{-7}$

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

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

 Step 2: Calculation of pH for Given Solution

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

Using Formula

$\begin{array}{l}\text{x}=\frac{\text{-b}\pm \sqrt{{\text{b}}^2-4\text{ac}}}{2a}\\ {\text{[H}}^{\text{+}}\text{]}=\frac{-(-7\times {10}^{-7})\pm \sqrt{{(-7\times {10}^{-7})}^2-4\left(1\right)(-1\times {10}^{-14})}}{2\left(1\right)}\\ {\text{[H}}^{\text{+}}\text{]}=7.14\times {10}^{-7}\end{array}$

${\text{pH=-log[H}}^{\text{+}}]=-\log (7.14\times {10}^{-7})=6.15$