Embibe Experts Solutions for Chapter: Equilibrium, Exercise 2: Exercise-2

What $\%$ of the carbon in the ${\mathrm{H}}_2{\mathrm{CO}}_3-{\mathrm{HCO}}_3^{-}$ buffer should be in the form of ${\mathrm{HCO}}_3^{-}$ so as to have a neutral solution? $\left({\mathrm{K}}_{\mathrm{a}}=4\times {10}^{-7}\right)$

Buffer capacity of a buffer solution is $\mathrm{x}$, the volume of $1 \mathrm{M} \mathrm{NaOH}$ added to $100 \mathrm{mL}$ of this solution if change the $\mathrm{pH}$ by $1$ is:

The total number of different kind of buffers obtained during the titration of ${\mathrm{H}}_3{\mathrm{PO}}_4$ with $\mathrm{NaOH}$ are :

The solubility product of ${\mathrm{Ag}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ at $25° \mathrm{C}$ is $1.29\times {10}^{-11}{\mathrm{mol}}^3{\mathrm{L}}^{-3}.$ A solution of ${\mathrm{K}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ containing
$0.152$ mole in $500 \mathrm{mL}$ water.is shaken with an excess of ${\mathrm{Ag}}_2{\mathrm{CO}}_3$ till the following equilibrium is reached:

${\mathrm{Ag}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)+{\mathrm{K}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Ag}}_2{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{s}\right)+{\mathrm{K}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)$.

At equilibrium, the solution contains $0.0358$ mole of ${\mathrm{K}}_2{\mathrm{CO}}_3$. Assuming the degree of dissociation of ${\mathrm{K}}_2{\mathrm{C}}_2{\mathrm{O}}_4$ and ${\mathrm{K}}_2{\mathrm{CO}}_3$ to be equal, if the solubility product of ${\mathrm{Ag}}_2{\mathrm{CO}}_3$ is $3.794\times {10}^{-\mathrm{x}} {\mathrm{mol}}^3{\mathrm{L}}^{-3}$, find $\mathrm{x}.$

$0.1$ millimole of ${\mathrm{CdSO}}_4$ are present in $10 \mathrm{mL}$ acid solution of $0.08 \mathrm{M} \mathrm{HCI}$. Now ${\mathrm{H}}_2\mathrm{S}$ is passed to precipitate all the ${\mathrm{Cd}}^{2+}$ ions. Find the $\mathrm{pOH}$ of the solution after filtering off precipitate, boiling off ${\mathrm{H}}_2\mathrm{S}$ and making the solution $100 \mathrm{mL}$ by adding ${\mathrm{H}}_2\mathrm{O}.$

$\mathrm{C} \mathrm{M}$ fluoroacetic acid solution was found to contain $\left[{\mathrm{H}}^{+}\right]=1.5\times {10}^{-3} \mathrm{M}. {\mathrm{K}}_{\mathrm{a}}$ of fluoroacetic acid $=2.5\times {10}^{-3}$. Then:

$50 \mathrm{mL}$ of $0.1 \mathrm{M} \mathrm{NaOH}$ is added to $60 \mathrm{mL}$ of $0.15 \mathrm{M} {\mathrm{H}}_3{\mathrm{PO}}_4$ solution (${\mathrm{K}}_1, {\mathrm{K}}_2$ and ${\mathrm{K}}_3$ for ${\mathrm{H}}_3{\mathrm{PO}}_4$ are ${10}^{-3}, {10}^{-}$ and ${10}^{-13}$ respectively). The $\mathrm{pH}$ of the mixture would be about $\left(\log 2=0.3\right)$

An acid-base indicator which is a weak acid has a ${\mathrm{pK}}_{\mathrm{In}}$ value $=5.45$. At what concentration ratio of sodium acetate to acetic acid would the indicator show a colour halfway between those of its acid and conjugate base forms? [${\mathrm{pK}}_{\mathrm{a}}$ of acetic acid $=4.75, \log 2=0.3$]