Question

Which of the following statements are correct at ${25}^{\circ }C?$

• A. $pOH=12$ for $0.01MNaOH$
• B. $pOH=12$ for ${10}^{-2}MHCl$
• C. $pH=3$ for ${10}^{-3}MHCl$
• D. $pOH=11$ for ${10}^{-3}MNaOH$

Answer 1

Answer: (B), (C)

The correct options are
B $pOH=12$ for ${10}^{-2}MHCl$
C $pH=3$ for ${10}^{-3}MHCl$

Since $HCl$ is a strong acid and $NaOH$ is a strong base, they both will dissociate completely into their ions . $[NaOH{]}_{initially}=[O{H}^{-}{]}_{eqb.}$
$[HCl{]}_{initially}=[H^{+}{]}_{eqb.}$

a.
$0.01MNaOH$
$\Rightarrow \left[O{H}^{-}\right]=0.01M$
So,
$pH=14-pOH$
$pH=14-(-log[O{H}^{-}\left]\right)$
$pH=14-(-log(0.01\left)\right)$
$pH=14-\left(2\right)=12$


b.
${10}^{-2}MHCl$
$\Rightarrow \left[H^{+}\right]={10}^{-2}M$
So,
$pH=-log\left[H^{+}\right])$
$pH=-log\left[{10}^{-2}\right])$
$pH=2pOH=14-pH=14-2=12$


c.
${10}^{-3}MHCl$
$\Rightarrow \left[H^{+}\right]={10}^{-3}M$
So,
$pH=-log\left[H^{+}\right]pH=-log\left({10}^{-3}\right)=3pH=3$

d.
${10}^{-3}MNaOH$
$\Rightarrow \left[O{H}^{-}\right]={10}^{-3}M$
So,
$pH=14-pOH$
$pH=14-(-log[O{H}^{-}\left]\right)$
$pH=14-(-log[{10}^{-3}\left]\right)$
$pH=14-3=11$

So, (b) and (c) are the correct answers.