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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
Question
  1. Predict the acidic, basic or neutral nature of the following salts:
$NaCN, KBr, NaNO_2, NH_4NO_3$
  1. What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K? (For calcium sulphate $K_{sp} is 9.1 \times 10^{-6}$).
  2. At $450K ; K_p = 2.0 \times 10^{10}bar^{-1}$ for the reaction at equilibrium:
​​​​​​​​​​​​​​$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$
What is $K_c$ at this temperature?
✓

Answer

$NaCN, NaNO_2$ - solutions are basic as they are salts of strong base and weak acid. (HCN and $HNO_2$ are weak acids and NaOH is strong base).
KBr - this solution is neutral as it is salt of strong acid HBr and strong base KOH.
$NH_4NO_3$ - its solution is acidic as it is a salt of strong acid $(HNO_3)$ and weak base $(NH_4OH)$.
  1. $\text{CaSO}_4(\text{s})\rightleftharpoons\text{Ca}^{2+}(\text{aq})+\text{SO}^{2-}_4(\text{aq});$
(mol. mass of $CaSO_4 = 136g/ mol$)
Let the solubility of $CaSO_4$ in mol/L is x.
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]=\text{x}^2$
$\text{x}=\sqrt{\text{K}_\text{sp}}=\sqrt{9.1\times10^{-6}}$
$=3.02\times10^{-3}\text{mol/L}$
$=3.02\times10^{-3}\times136\text{g/L}$
$=0.411\text{g/L};$
For dissolving $0.411g$ of $CaSO_4$, water required is 1L
For dissolving 1g of $CaSO_4$, water required $=\frac{1}{0.411}\text{L}=2.43\text{L}$
  1. For the reaction:
$2\text{SO}_2(\text{g})+\text{O}_2\text{(g)}\rightleftharpoons2\text{SO}_3(\text{g})$
$\Delta\text{n}_\text{g}=2-3=-1$
$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$
$\text{K}_\text{c}=\text{K}_\text{p}(\text{RT})^{-\Delta\text{n}}$
For this reaction
$\text{K}_\text{c}=\text{K}_\text{p}\text{RT}=(2\times10^{10}\text{bar}^{-1}) $
$(0.083\text{L bar K}^{-1}\text{mol}^{-1})\times450\text{K}$
$=7.48\times10^{11}\text{L mol}^{-1}$

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