Question

Calculate the pH and {OH-} of solution with {H+}=4.2x10-9

Answer 1

Calculate the pH and $\{\mathrm{OH - }\}$ of solution with $\{\mathsf{H}^{+}\} = 4.2\times 10^{-9}$

Solution

Find the pH of this solution:

$$p H = - l g \{H ^ {+} \} = - \lg (4. 2 \cdot 1 0 ^ {- 9}) = 8. 3 7 7$$

Water molecules auto-dissociate into $\{\mathsf{H}^{+}\}$ and $\{\mathsf{OH}^{-}\}$ ions in the following equilibrium:

$$H _ {2} O \leftrightarrow H ^ {+} + O H ^ {-};$$

The concentration product of these ions in water solution is constant:

$$\{H ^ {+} \} \cdot \{O H ^ {-} \} = 1 \cdot 1 0 ^ {- 1 4} \mathrm {M} ^ {2};$$

So, if $\{H^{+}\} = 4.2\cdot 10^{-9}$ M:

$$\{O H ^ {-} \} = \frac {1 \cdot 1 0 ^ {- 1 4}}{\{H ^ {+} \}} = \frac {1 \cdot 1 0 ^ {- 1 4}}{4 . 2 \cdot 1 0 ^ {- 9}} = 0. 2 3 8 \cdot 1 0 ^ {- 5} = 2. 3 8 \cdot 1 0 ^ {- 6} \mathrm {M};$$

Answer: the pH of solution is 8.377; $\{OH^{-}\} = 2.38\cdot 10^{-6}$ M.