MCQ
What will be the $pH$ of a solution formed by mixing $40\,\,ml$ of $0.10\,\,M\,\,HCl$ with $10\,\,ml$ of $0.45\,\,M\,\,NaOH$
- ✓$12$
- B$10$
- C$8$
- D$6$
M.eq. of $0.45$ $M\,\,NaOH = \frac{{0.45 \times 10}}{{1000}} = 0.0045\,M$
Now left $[O{H^ - }] = 0.0045 - 0.004$$ = 5 \times {10^{ - 4}}$ $M$
Total volume $ = 50 \,ml.$
$[O{H^ - }] = \frac{{5 \times {{10}^{ - 4}}}}{{50}} \times 1000$; $[O{H^ - }] = 1 \times {10^{ - 2}}$
$pOH = 2 ; pH = 14 -pOH = 12.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
......$kJ/mole$