Question

Show that the hydroxide-ion concentration in an aqueous solution is 1x10^-4 M when the hydronium-ion concentration is 1x10^-10 M. Recall that 10^a x 10^b x =10^(a+B)

Answer 1

Show that the hydroxide-ion concentration in an aqueous solution is $1 \times 10^{\wedge} - 4$ M when the hydronium-ion concentration is $1 \times 10^{\wedge} - 10$ M. Recall that $10^{\wedge} a \times 10^{\wedge} b \times 10^{\wedge} (a + B)$

Solution

Water molecules auto-dissociate into hydronium and hydroxide ions in the following equilibrium:

$$2 H _ {2} O \leftrightarrow H _ {3} O ^ {+} + O H ^ {-};$$

The concentration product of hydronium and hydroxide ions in water solution is constant:

$$\left[ H _ {3} O ^ {+} \right] \cdot \left[ O H ^ {-} \right] = 1 \cdot 1 0 ^ {- 1 4} \mathrm {M} ^ {2};$$

So, if the hydronium-ion concentration is $[H_3O^+] = 1 \cdot 10^{-10} \mathrm{M}$ , then hydroxide-ion concentration is:

$$\left[ O H ^ {-} \right] = 1 \cdot \frac {1 0 ^ {- 1 4}}{\left[ H _ {3} O ^ {+} \right]} = 1 \cdot \frac {1 0 ^ {- 1 4}}{1 0 ^ {- 1 0}} = 1 \cdot 1 0 ^ {- 4} \mathrm {M};$$

Answer: in this case the hydroxide-ion concentration in an aqueous solution is $10^{-4}\mathrm{M}$ when the hydronium-ion concentration is $10^{-10}\mathrm{M}$ , because the concentration product of hydronium and hydroxide ions in water solution is constant $(1\cdot 10^{-14}\mathrm{M}^2)$ .