MCQ
$200\; mL$ of $0.01\; M\; HCl$ is mixed with $400 \;mL$ of $0.01\; M\; H _{2} SO _{4}$. The $pH$ of the mixture is
- A$1.14$
- ✓$1.78$
- C$2.34$
- D$3.02$
$\left[ H ^{+}\right]=\frac{(0.01 \times 200)+(0.01 \times 2 \times 400)}{600}$
$=\frac{2+8}{600}=\frac{10}{600}=\frac{1}{60}$
$pH =-\log [\frac{1}{60}]$
$=1.78$
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(Given : Energy of the electron in the first shell of the hydrogen atom is $-2.2 \times 10^{-18} \,J$; $h =6.63 \times 10^{-34}\, Js$ and $c =3 \times 10^{8} \,ms ^{-1}$ )
$\frac{1}{2}C{l_2}(g)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl(g)\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}$ $C{l^ - }(g)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }(aq)$
(using the data,
${{\Delta _{diss}}H_{C{l_2}}^\Theta } = 240\,kJ\,mol^{-1}, {{\Delta _{eg}}H_{C{l}}^\Theta }= -349 \,kJ\,mol^{-1},$${{\Delta _{Hyd}}H_{C{l}}^\Theta }= -381 \,kJ\,mol^{-1}$ ) will be ............. $\mathrm{kJ\,mol}^{-1}$