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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
  1. Define solubility product. Write solubility product expression in terms of molar solubility for $FeCl_3$
  2. What is the effect of temperature on solubility of gases in liquids?
  3. Equilibrium constant for the reaction is $4.0$. What will be the equilibrium constant for the reverse reaction.
  4. Calculate the pH of $10^{-8}M$ HCl solution.
✓

Answer

  1. Solubility product is defined as the product of molar concentration of ions raised to the power the number of ions formed per formula of the compound.
$\text{FeCl}_3(\text{s})\rightleftharpoons\text{Fe}^{3+}+3\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ '\text{s}' \ \ \ \ \ \ \ \ \ \ 3\text{s}$
$\text{K}_{\text{sp}}=[\text{Fe}^{3+}][\text{Cl}^-]^3$
$\text{K}_\text{sp}=(\text{s})(3\text{s})^3,$
where 's' $mol L^{-1}$ is solubility.
$\Rightarrow\text{K}_\text{sp}= 27\text{s}^4$
$\Rightarrow\text{s}=4\sqrt{\frac{\text{K}_\text{sp}}{27}}$
  1. Solubility of gases in liquids decreases with increase in temperature because force of attraction between gas and liquid decreases at high temperature.
  2. K for the reaction = 4
$\therefore \text{K}'$ for reverse reaction $=\frac{1}{4}=0.25$ [$\because \text{K}'=\frac{1}{ \text{K}}$ for reverse reaction]
  1. pH of $10^{-8}M$ HCl solution.
$\text{HCl}\xrightarrow{\ \ \ \ \ \ }\text{H}^++\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{10^{-8}\text{M}}$
$\text{H}_2\text{O}\rightleftharpoons\text{H}^++\text{OH}^-$
$\text{K}_\text{w}=1\times10^{-14}$
$\Rightarrow [\text{H}^+][\text{OH}^-]=10^{-14}$
$\because[\text{H}^+]=[\text{OH}]$
$\therefore [\text{H}^+]^2=10^{-14}$
$\Rightarrow [\text{H}^+]=10^{-7}\text{mol L}^{-1}$
Total concentration of
$[\text{H}^+]=(10^{-8}+10^{-7})$
$=10^{-7}(1+0.1)$
$=1.1\times10^{-7}$
$\therefore \text{pH}=-\log[\text{H}^+]$
$=-\log1.1\times10^{-7}$
$=-\log1.1-\log10^{-7}$
$\Rightarrow \text{pH}=-0.0454+7.000$
$=6.9546$

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