Chapter 8: Q32P (page 186)

(a) Ion Pairing . As in problem $8 - 30$ , find the concentration, ionic strength, and ion pair fraction in $0.025 F {MgSO}_{4}$
(b) Two possibly important reactions that we did not consider are acid hydrolysis of ${Mg}^{+} ({Mg}^{2 +} + H_{2} O ֏ {MgOH}^{+} + H^{+})$ and base hydrolysis of ${SO}^{2 -}_{4}$ . Write these two reactions and find their equilibrium constants in Appendices I and G. With the assumed pH near $7.20$ and neglecting activity coefficient, show that both reactions are negligible.

Short Answer

Expert verified

Thus the Ionic Strength and Ion Pair Fraction of $0.025 F {MgSO}_{4}$ is $0.06463 M$ and $35.4 %$ and the equilibrium constants are $6 \times 10^{- 7} M and 2 \times 10^{- 7} M$

Step by step solution

Step 1:Calculating the Ionic Strength and Ion Pair Fraction.

This is also a task that need to be performed in the Excel. And the values will be,

$[{Mg}^{2 +}] = [{SO}^{2 -}_{4}] = 0.01616 M [{MgSO}^{-}_{4}] = 0.008844 M$

Ionic Strength $= 0.06463 M$

Ion Pair Fraction $= 35.4 %$

Step 2:Calculating the equilibrium Constants.

In the part (B) , we need to find the equilibrium constant.

${Mg}^{2 +} + {OH}^{-} ֏ {MgOH}^{+} + K_{1} = 10^{2.6} [{MgOH}^{+}] {֏K}_{1} [{Mg}^{2 +}] [{OH}^{-}] = 10^{2.6} \times 0.016 \times 10^{- 7} = 6 \times 10^{- 7} M$

This is negligible in comparison to $[{Mg}^{2 +}] = 0.016 M$

${SO}_{4}^{2 -} + H_{2} O ֏ {HSO}^{-}_{4} + {OH}^{-} {pK}_{b} = 12.01 [{HSO}^{-}_{4}] ֏ K_{b} [{SO}^{2 -}_{4}] / [{OH}^{-}] = 10^{- 12.01} \times \frac{0.016}{10^{- 7}} = 2 \times 10^{- 7} M$

This is negligible in comparison to $[{SO}_{4}^{2 -}] = 0.016 M$

Thus the Ionic Strength and Ion Pair Fraction of $0.025 F {MgSO}_{4} is 0.06463 M$ and $35.4 %$ and the equilibrium constants are $6 \times 10^{- 7} M and 2 \times 10^{- 7} M$

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Short Answer

Step by step solution

Step 1:Calculating the Ionic Strength and Ion Pair Fraction.

Step 2:Calculating the equilibrium Constants.

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