MCQ
$20\,ml$ of $0.5\,N\,HCl$ and $35\,ml$ of $0.1\,N\,NaOH$ are mixed. The resulting solution will
- ABe neutral
- BBe basic
- ✓Turn phenolphthalein solution pink
- DTurn methyl orange red
$0.5\,N \Rightarrow 1000\,ml\,0.5\,mole\,HCl$ is present in $20\,ml$
$ = \frac{{20 \times 0.5}}{{1000}} = 1.0 \times {10^{ - 2}}$
$(ii)$ $35\, ml$ of $0.1\, N \,NaOH$
$0.1\,N \Rightarrow 1000\,ml$ of $0.1 \,mole $ $NaOH$ is $35 $ $ml$
$ = \frac{{35 \times 0.1}}{{1000}} = 0.35 \times {10^{ - 2}}$
Total $= 20 + 35 = 55\, ml.$
$ \Rightarrow $ $(1.0-0.35)10^{-2}=0.65 \times 10^{-2}$ $mole$ $HCl$
$HCl = H^++Cl^-$
$ \Rightarrow $ $[HCl] = [H^+]+[Cl^-]$
$55 \,ml $ contains $ 0.65 \times 10^{-2} \,mole $ of $H^+ $ ions
$1000\,ml - \frac{{0.65 \times {{10}^{ - 2}} \times {{10}^3}}}{{55}} = \frac{{6.5}}{{55}}$
$pH = - \log [{H^ + }] = - \log (6.5/55)$
$ = \log 55 - \log 6.5 = 0.92$
Due to acidic nature of solutions the colour of phenolphthalein becomes pink.
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$[A]$ With increase in temperature, the value of $K$ for exothermic reaction decreases because entropy change of the system is positive
$[B]$ With increase in temperature, the value of $K$ for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases
$[C]$ With increase in temperature, the value of $K$ for endothermic reaction increases because the entropy change of the system is negative
$[D]$ With increase in temperature, the value of $K$ for exothermic reaction decreases because favourable change in entropy of the surrounding decreases
$\mathop {{H_3}C\mathop C\limits^ \ominus {H_2}}\limits_{(i)} \,,$$\mathop {{H_2}C = \mathop C\limits^ \ominus H}\limits_{(ii)} $ and $\mathop {H - C \equiv \mathop C\limits^ \ominus }\limits_{(iii)} $
is in the order of