Name: Applications of Solubility Product
Uploaded: 2019-07-19
Duration: 8 min 36 s
Description: Solubility product helps determine the solubility of sparingly soluble salts and the condition of salt precipitation. Learn more in this video.

Question

Calculate the change in $pH$ when $4 g$ of solid $NaOH$ and $10 mmol$ of $H_{2} {SO}_{4}$ are added to a solution of volume $5$ $litre$ , which was prepared by mixing $20 mmol$ of $HCl$ . $40 mmol$ of $H_{2} {SO}_{4}$ and $2 g$ of $NaOH$ and sufficient water. Ignore the volume change.

$(\log 2 = 0.3, \log 3 = 0.48)$

Accepted Answer

Before mixing

Total no of $H^{+}$ ions $NaOH$ $20 \times 10^{- 3} (from HCl) + 2 \times 40 \times 10^{- 3} (from H_{2} {SO}_{4}) = 0.1 mol$

Total no of ${OH}^{-}$ ions $=$ $2 \div 40 = 0.05 mol$

Now hydroxide ions are lesser than hydrogen ions. Hence the solution is acidic

Hence no of $H^{+}$ ions $=$ $0.1 - 0.05 = 0.05 mol$

$[H^{+}] = \frac{0.05}{5} = 0.01 {pH}_{1} = - \log (0.01) = 2$

After mixing

Total no of $H^{+}$ ions $10 mmol$ $20 \times 10^{- 3} (from HCl) + 40 \times 2 \times 10^{- 3} (from H_{2} {SO}_{4}) + 10 \times 2 \times 10^{- 3} (from H_{2} {SO}_{4}) = 0.12 mol$

Total no of ${OH}^{-}$ ions $=$ $\frac{4 + 2}{40} mol = 0.15 mol$

Now no of hydroxide ions are more than hydrogen ions, hence the solution is basic

${[{OH}^{-}]}_{net} = \frac{0.15 - 0.12}{5} = 6 \times 10^{- 3} pOH = 3 - \log 6 = 2.22 pH = 14 - 2.22 = 11.78$

Now Change in
$pH = 11.78 - 2 = 9.78$

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !

Calculate the change in pH when 4 g of solid NaOH and 10 mmol of H2SO4 are added to a solution of volume 5 litre, which was prepared by mixing 20 mmol of HCl. 40 mmol of H2SO4 and 2 g of NaOH and sufficient water. Ignore the volume change. log 2=0.3, log3=0.48

Important Questions on Ionic Equilibrium

Learn and Prepare for any exam you want !