- ✓$20$
- B$16$
- C$24$
- D$10$
or $7.4=6.1+\log \frac{\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}$
or $\frac{\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}=20$
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${\text{2PQ}} \rightleftharpoons {{\text{P}}_2}{\text{ + }}{{\text{Q}}_2}\,;\,\,\,\,{{\text{K}}_1}{\text{ = 2}}{\text{.5 }} \times \,{\text{1}}{{\text{0}}^5}\,$
${\text{PQ + }}\frac{1}{2}{R_2} \rightleftharpoons PQR\,;\,\,\,\,{{\text{K}}_2}{\text{ = 5 }} \times \,{\text{1}}{{\text{0}}^{ - 3}}\,$
The value of equilibrium constant for reaction
$\frac{1}{2}{{\text{P}}_2}{\text{ + }}\frac{1}{2}{{\text{Q}}_2}{\text{ + }}\frac{1}{2}{R_2} \rightleftharpoons PQR\,$ is

$MnCl _2+ K _2 S _2 O _8+ H _2 O \longrightarrow KMnO _4+ H _2 SO _4+ HCl$ (equation not balanced).
Few drops of concentrated $HCl$ were added to this solution and gently warmed. Further, oxalic acid ( $225 mg$ ) was added in portions till the colour of the permanganate ion disappeared. The quantity of $MnCl _2$ (in $mg$ ) present in the initial solution is. . . . . . . . . (Atomic weights in $g mol ^{-1}: Mn =55, Cl =35.5$ )