Question

Calulate the pH of the following solutions :
(i) $2.21g$ of $T\ell OH$ dissolved in water to give $2$ litre of solution. (Assume $T\ell OH$ to be a strong base)
(ii) $0.49$ w/v $H_{2}S{O}_4$ solution
(iii) $\frac{M}{1000}Sr{\left(OH\right)}_2$ solution is diluted to quadruple volume.
(iv) $1mL$ of $12MHCl$ is diluted with water to obtain $1litre$ of solution.

Answer 1

(i) $2.21g$ of $TlOH⟶$

$TlOH+H_{2}O⟶2Lso{l}^n$

$2.21g$

molecular weight of $TlOH=221.39$

no. of moles of $TlOH=\frac{2.21}{221.39}$

concentration of $TlOH$ is $2L$ solution=$\frac{2.21}{221.39}\times \frac{1}{2}$

$=0.00499\sim 0.005M$

$pOH=-\log \left[O{H}^{+}\right]$

$=-\log \left[0.005\right]$

$2.301$

$pH=14-2.301=11.699$

(ii) $0.49w/v$ $H_{2}S{O}_{4}so{l}^n$

measuring $0.49g$ present in $100mL$ of solution

no. of $g-eq$ of $H_{2}S{O}_4=weight/g-eq$ of $H_{2}S{O}_4$

$=\frac{0.49}{49}$ moles

$=0.01$ moles

normality=$(Equivalent\times 1000)/Volumeofsolution$

$=\frac{0.01\times 1000}{100}$

$=0.1N$

n-factor of $$H_2SO_4$is$2$

So, $0.1=2\times M$

$M=0.05$

$pH=-\log \left(0.05\right)=1.301$

(iii) $\frac{M}{1000}Sr(OH{)}_2$ solution is diluted to quadruple volume.

concentration decreases by $\frac{1}{4}$ times as the solution is diluted to quadruple volume.

$pOH=-\log \left[O{H}^{-}\right]=-\log \left[\frac{n}{v}\right]$

$=-\log \left[\frac{n}{4v}\right]$

$pOH=-log\left[\frac{{10}^{-3}}{4}\right]$

$=0.602+3=3.602$

$pH=14-3.602=10.398$

(iv) $1mL$ of $12M$ $HCl$ is diluted with water to obtain $1L$ of solution.

$M_1{V}_1=M_2{V}_2$

$12\times 1mL=M_2\times 1000$

$M_2=\frac{12}{1000}=12\times {10}^{-3}M$

$\left[H^{+}\right]$=$12\times {10}^{-3}M$

$pH=-\log \left[H^{+}\right]$

$=\log [12\times {10}^{-3}]$

$=-\log 12-\log {10}^{-3}$

$=-1.079+3=1.921$