Question

The $pH$ of the solution obtained by mixing $100\, ml$ of a solution of $pH = 3$ with $400\, ml$ of a solution of $pH = 4$ is

• A. 3 - log 2.8
• B. 4 - log 2.8
• C. 7 - log 2.8
• D. 5 - log 2.8

Answer 1

Answer: (B)

4 - log 2.8

Answer 2

For solution I, $pH =3$
$\Rightarrow\left[ H ^{+}\right]=10^{-3}$
$\Rightarrow$ Concentration of solution $M _{1}=10^{-3} M$
$V_{1}=100\, mL$

For solution $II,\, pH =4$
$\Rightarrow\left[ H ^{+}\right]=10^{-4}$
$\Rightarrow$ Concentration of solution $M _{1}=10^{-4} \,M$
$V_{2}=400\, mL$

Concentration of resulting solution

$M=\frac{M_{1} V_{1}+M_{2} V_{2}}{V_{1}+V_{2}}$
$=\frac{10^{-3} \times 100+10^{-4} \times 400}{100+400}$
$=\frac{0.14}{500}$
$M=0.00028=2.8 \times 10^{-4}$
$\therefore\left[H^{+}\right]=2.8 \times 10^{-4}$
$\therefore pH$ of resulting solution

$=-\log \left[ H ^{+}\right]$
$=-\log \left(2.8 \times 10^{-4}\right)$
$=-\log 2.8-\log 10^{-4}$
$=4-\log 2.8$