Question

What are the concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2?

Answer 1

Answer: $[H^+]=6.3\times 10^{-11}$  and $[OH^-]=1.58\times 10^{-4}$

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

$pH=-\log [H^+]$

$pOH=-log[OH^-]$

$pH+pOH=14$

Given pH = 10.2

$10.2=-\log [H^+]$

$[H^+]=6.3\times 10^{-11}$

$pOH=14-10.2=3.8$

$3.8=-log[OH^-]$

$[OH^-]=1.58\times 10^{-4}$

Answer 2

Answer : The concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2 are, $1.58\times 10^{-4}$ and $6.3\times 10^{-11}$ respectively.

Explanation : Given,

pH = 10.2

pH : It is defined as the negative logarithm of the hydrogen ion concentration.

First we have to calculate the hydrogen ion concentration $(H^+)$

$pH=-\log [H^+]$

Now put the value of pH in this formula, we get the hydrogen ion concentration.

$10.2=-\log [H^+]$

$[H^+]=6.3\times 10^{-11}$

Now we have to calculate the pOH of the solution.

$pH+pOH=14$

Now put the value of pH, we get the value of pOH.

$10.2+pOH=14$

$pOH=14-10.2$

$pOH=3.8$

Now we have to calculate the hydroxide ion concentration $(OH^-)$

$pOH=-\log [OH^-]$

Now put the value of pOH in this formula, we get the hydroxide ion concentration.

$3.8=-\log [OH^-]$

$[OH^-]=1.58\times 10^{-4}$

Therefore, the concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2 are, $1.58\times 10^{-4}$ and $6.3\times 10^{-11}$ respectively.