Question

Which of the following solutions will have pH close to 1.0?
log 2 = 0.3010

• A. $100mLof\left(M/10\right)HCl+100mLof\left(M/10\right)NaOH$
• B. $10mLof\left(M/10\right)HCl+90mLof\left(M/10\right)NaOH$
• C. $75mLof\left(M/5\right)HCl+25mLof\left(M/5\right)NaOH$
• D. None of the above

Answer 1

The correct option is C $75mLof\left(M/5\right)HCl+25mLof\left(M/5\right)NaOH$

(a) 100 mL (M/10) HCl will completely neutralise 100 mL (M/10) NaOH and the solution will be neutral.

(b) After neutralisation resultant solution will be basic due to presence of excess of NaOH.
$\text{M.eq. of HCl}=10\times \frac{1}{10}=1\text{m.eq}$
M.eq. of NaOH $=90\times \frac{1}{10}=9$ m.eq
$\therefore$ M. eq. of NaOH left = 9-1=8
$\therefore \left[OH\right]=\frac{8}{100}$
$\therefore pOH=-log\left[O{H}^{-}\right]=-log\left[\frac{8}{100}\right]$
$\therefore pOH=-log\left[\frac{8}{100}\right]=-log8+log100$
$\therefore pOH=1.09$
$pH+pOH=14pH=12.91$

(c) $\text{M.eq. of HCl}=75\times \frac{1}{5}=15\text{m.eq}$
M.eq. of NaOH $=25\times \frac{1}{5}=5$ m.eq
$\therefore$ M. eq. of HCl left = 15 - 5 = 10
$\therefore \left[HCl\right]=\frac{10}{100}=\frac{1}{10}$
$\therefore pH=-log\left[H^{+}\right]=-log\left[\frac{1}{10}\right]=1$