MCQ
By adding $20\,ml$ $0.1\,N\,HCl$ to $20\,ml$ $0.001\,N\,$ $KOH,$ the $ pH$ of the obtained solution will be
- A$2$
- ✓$1.3$
- C$0$
- D$7$
$20\,ml. $ of $0.001\, KOH = $ $\frac{{0.001}}{{1000}} \times 20\,gm$ eq.
$= 2 ×$ ${10^{ - 5}}\,g$ eq.
$\therefore $ $HCl $ left unneutralised = $2({10^{ - 3}} - {10^{ - 5}})$
$ = 2 \times {10^{ - 3}}(1 - 0.01)$ = $2 \times 0.99 \times {10^{ - 3}}$$ = 1.98 \times {10^{ - 3}}\,g\,eq.$
Volume of solution $= 40 \,ml.$ $ [HCl] =$ $\frac{{1.98 \times {{10}^{ - 3}}}}{{40}} \times 1000M$ = $4.95 \times {10^{ - 2}}$
$\therefore $ $pH = $ $2 - $ $log \,4.95 = 2 -0.7 = 1.3.$
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$(1)$ $NaN{H_2};\mathop {C{H_3}}\limits^{{\text{a}}\,\,\,\,\,\,\,} C{H_2}Br$:${H_2},{\text{ }}\mathop {{\text{(one mole)}}}\limits^{\text{b}} {\text{ }}Pd{\text{ or }}Ni$
$(2)$ $NaN{H_2};C{H_3}C{H_2}C{H_2}Br$:${H_2}{\text{ (two moles) }}Pd{\text{ or }}Ni$
$(3)$ $NaN{H_2};C{H_3}C{H_2}C{H_2}Br$:${H_2},{\text{ (one mole) }}Pd{\text{ or }}Ni$
$(4)$ $NaN{H_2};C{H_3}C{H_2}C{H_2}Br$:$B{H_3},{H_2}{O_2},O{H^ - }$
$(A)$ Homolytic bond cleavage
$(B)$ Heterolytic bond cleavage
$(C)$ Free radical formation
$(D)$ Primary free radical
$(E)$ Secondary free radical
Choose the correct answer from the options given below: