Question

$100mL$ of $0.01M$ solution of $NaOH$ is diluted to $1{dm}^3$. What is the $pH$ of the diluted solution?

• A. 12
• B. 11
• C. 2
• D. 3

Answer 1

The correct option is B 11

(i) $pOH=-\log \left[{OH}^{-}\right]$
(ii) $pH+pOH=14$
$\because 100mL0.01MNaOH$ solution is diluted to $1{dm}^3$ (ie, 10 times diluted)
$\therefore$ Resultant solution of $NaOH=0.001M$
$NaOH⟶{Na}^{+}+{OH}^{-}$
$\therefore \left[{OH}^{-}\right]=0.001={10}^{-3}$
$pOH=-\log \left[{OH}^{-}\right]$
$=-\log {10}^{-3}=3$
$pH+pOH=14$
$\therefore pH=14-pOH=14-3=11$