Question

A solution has a [OH−] of 1 × 10−12 M. What is the [H3O+] of the solution?

Answer 1

Answer:  1 × 10−2 M

Explanation:  Kw is known as the constant of water dissociation and its value is found to be $10^{-14}$  . It is the product of the concentration of  [H3O+] and [OH -] ions .

Dissociation of water can be represented by -

         $2 H_{2} O (aq)\rightleftharpoons H_{3}O^{+}(aq) + OH^{-}(aq)$

Kw = [H3O+]  [OH -]

As we know that pKw = - log Kw,

Thus pKw = 14

pOH = - log OH- = - log (10^{-12}) = 12

and  pKw = pH + pOH

           14   = pH + 12

pH = 14 - 12 = 2

pH = -log H3O+

2 =   -log H3O+

Therefore, [H3O+] = 1 × 10−2 M

Answer 2

Answer: The correct answer is Option B.

Explanation:

To calculate the pH of the solution, we need to determine pOH of the solution. To calculate pOH of the solution, we use the equation:

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

We are given:

$[OH^-]=1\times 10^{-12}M$

Putting values in above equation, we get:

$pOH=-\log(1\times 10^{-12})\\\\pOH=12$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-12=2$

pH is defined as negative logarithm of hydrogen ion or hydronium ion concentration. It is basically defined as the power of hydronium ions in a solution.

$pH=-\log[H_3O^+]$

pH of the solution = 2

Putting values in above equation, we get:

$2=-\log[H_3O^+]$

$[H_3O^+]=antilog(-2)$

$[H_3O^+]=1\times 10^{-2}$

Hence, the correct answer is Option B.