Question

Calculate the $pH$ of the following solutions:

a) 2 g of $TlOH$ dissolved in water to give 2 litre of solution.
b) 0.3 g of $Ca(OH{)}_2$ dissolved in water to give 500 mL of solution.
c) 0.3 g of $NaOH$ dissolved in water to give 200 mL of solution.
d) 1 mL of 13.6 M $HCl$ is diluted with water to give 1 litre of solution.

Answer 1

(a) For 2g of $TlOH$ dissolved in water to give 2 L of solution:

$\left[TlOH\right]=\left[O{H}^{-}\right]=\frac{2\times 1}{2\times 221}=\frac{1}{221}M$

$pOH=-log[OH{]}^{-}=-log\frac{1}{221}=2.35$

$pH=14-pOH=14-2.35=11.65$

(b) For 0.3 g of $Ca(OH{)}_2$ dissolved in water to give 500 mL of solution:

$\left[O{H}^{-}\right]=2\left[Ca\right(OH{)}_2]=2(0.3\times \frac{1000}{500})=1.2M$

$pOH=-log\left[O{H}^{-}\right]=-log1.2=1.79$

$pH=14-pOH=14-1.79=12.21$
(c) For 0.3 g of $NaOH$ dissolved in water to give 200 mL of solution:

$\left[O{H}^{-}\right]=\left[NaOH\right]=0.3\times \frac{1000}{200}=1.5M$

$pOH=-log\left[O{H}^{-}\right]=-log1.5=1.43$

$pH=14-pOH=14-1.43=12.57$
(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

The molarity of HCl solution after dilution is $\frac{13.6\times 1}{1000}=0.0136M$.

It is equal to hydrogen ion concentration.

$pH=-log\left[H^{+}\right]=-log0.0136=1.87$