$(b)\,\mathop {BaS{O_4} \downarrow }\limits_{(InSo\operatorname{lub} le)} + \mathop {2NaOH}\limits_{(So\operatorname{lub} le)} \xrightarrow{{Aqueous}}\mathop {Ba{{(OH)}_2}}\limits_{(So\operatorname{lub} le)} + \mathop {N{a_2}S{O_4}}\limits_{(So\operatorname{lub} le)} $
$(c)\,\mathop {Pb{{(N{O_3})}_2}}\limits_{(So\operatorname{lub} le)} + \mathop {2C{H_3}COONa}\limits_{(So\operatorname{lub} le)} \xrightarrow{{Aqueous}}\mathop {Pb{{(OAc)}_2}}\limits_{(So\operatorname{lub} le)} + \mathop {2NaN{O_3}}\limits_{(So\operatorname{lub} le)} $
$(d)\,\mathop {N{a_2}Cr{O_7}}\limits_{(So\operatorname{lub} le)} + \mathop {BaC{l_2}}\limits_{(So\operatorname{lub} le)} \xrightarrow{{Aqueous}}\mathop {BaCr{O_4}}\limits_{(InSo\operatorname{lub} le)} \downarrow + \mathop {2NaCl}\limits_{(So\operatorname{lub} le)} $
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[Atomic mass $: \mathrm{Na}=23.0, \mathrm{O}=16.0, \mathrm{H}=1.0]$
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}} \\
|
\end{array}} \\
{C{H_3} - C - C{H_3}} \\
| \\
H
\end{array}{\mkern 1mu} $ $\mathop {\xrightarrow{{C{H_3}OBr}}}\limits_{C{H_3}OH} $
$C{O_2} + 2{H_2}O\, \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $
for which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^- ]$ to $[CO_2]$ would be :-
