Question

What are the $pH$ values of the following solutions at ${25}^{o}C$?
(i) 0.001 M of $HCl$
(ii) 0.05 M of $H_{2}S{O}_4$
(iii) 0.01 M of $NaOH$

• A. (i) $pH=1$, (ii) $pH=12$, (iii) $pH=3$
• B. (i) $pH=3$, (ii) $pH=1$, (iii) $pH=12$
• C. (i) $pH=1$, (ii) $pH=3$, (iii) $pH=12$
• D. None of these

Answer 1

The correct option is B (i) $pH=3$, (ii) $pH=1$, (iii) $pH=12$

$\left(i\right)$ Given Concentration of $HCl$ which is nothing but $\left[HCl\right]=\left[{H_{3}O}^{+}\right]=0.001M$

We know that $pH=-\log \left[{H_{3}O}^{+}\right]=-\log (0.001)=3$

$\left(ii\right)$ Given Concentrtion of $H_{2}S{O}_4=0.05M$

Here $\left[{H_{3}O}^{+}\right]=2\times 0.05M$ (as it has two Hydrogens)

$\left[{H_{3}O}^{+}\right]=0.1M$

Now $pH=-\log \left[{H_{3}O}^{+}\right]=-\log (0.1)=1$

$\left(iii\right)$ Given Concentration of $NaOH$ which is $0.01M$

We know that $pOH=-\log \left[{OH}^{-}\right]=-\log \left(0.01\right)=2$

We also know that $pH+pOH=14pH=14-pOHpH=14-2pH=12$