# Question 31148

Feb 2, 2017

The given $N a O H$ solution is 5%w/w#

This means 100g solution contains 5g $N a O H$ and $95 g$ water.

The density of solution is not known. So the volume of solution can not be found out.

If we assume that the solution being dilute the volume of solution is approximately equal to the volume of solvent water, then we can proceed to calculate the pH of the solution. Taking density of water $1 g \text{/cc}$ the volume of water as well as solution will be $95 c {m}^{3} = 95 \times {10}^{-} 3 L$

The amount of $N a O H$ in solution

$= 5 g = \frac{5 g}{40 g \text{/mol}} = \frac{1}{8} = 0.125 m o l$

So the concentrantion of $N a O H$
$= \frac{0.125}{95 \times {10}^{-} 3} M = \frac{25}{19} M$

So the concentration of $O {H}^{\text{-}} = \frac{25}{19} M$

So $p H = p {K}_{w} - p O H$

$= 14 - {\log}_{10} = 14 - {\log}_{10} \left[O {H}^{\text{-}}\right]$

$= 14 - {\log}_{10} \left(\frac{25}{19}\right) = 13.88$