Solve this following
Question:

The $\mathrm{pH}$ of a solution obtained by mixing $50 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{HCl}$ and $30 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{NaOH}$ is $\mathrm{x} \times 10^{-4}$. The value of $\mathrm{x}$ is (Nearest integer)

$[\log 2.5=0.3979]$

Solution:

$\mathrm{HCl}$ (aq.) $+\mathrm{NaOH}$ (aq.) $\rightarrow \mathrm{NaCl}$ (aq.) $+\mathrm{H}_{2} \mathrm{O}(\ell)$

$50 \mathrm{ml}, 1 \mathrm{M} \quad 30 \mathrm{ml}, 1 \mathrm{M}$

$\mathrm{t}=0 \quad 50 \mathrm{~mm} \quad 30 \mathrm{~mm}$

$\mathrm{t}=\infty \quad 20 \mathrm{~mm}$

$[\mathrm{HCl}]=\frac{20}{80}=\frac{1}{4} \mathrm{M}=2.5 \times 10^{-1} \mathrm{M}$

$\mathrm{pH}=-\log 2.15 \times 10^{-1}=1-0.3979=0.6021$

$\mathrm{pH}=6021 \times 10^{-4}$

 

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