Question

When an aqueous solution at room temperature is analyzed, the [H+] is found to be 2.0 × 10−3 M. What is the [OH−]?

Answer 1

Answer : The correct option is, (A) $5.0\times 10^{-12}M$

Explanation:

First we have to calculate the pH.

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

$pH=-\log (2.0\times 10^{-3})$

$pH=2.69$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-2.69=11.31$

Now we have to calculate the $OH^-$ concentration.

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

$11.31=-\log [OH^-]$

$[OH^-]=5.0\times 10^{-12}M$

Therefore, the $OH^-$ concentration is, $5.0\times 10^{-12}M$

Answer 2

The answer would be A. 5.0x10-12m