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  4. 4 ml of HCl solution of pH = 2 is mixed with 6 ml of NaOH solution of pH = 12. What would be the final pH of solution? (log 2 = 0.3)

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Q. 4 ml of HCl solution of pH = 2 is mixed with 6 ml of NaOH solution of pH = 12. What would be the final pH of solution? (log 2 = 0.3)

NTA AbhyasNTA Abhyas 2020Equilibrium

Solution:

$pH=2, \, \left[H C l\right]=\left(10\right)^{- 2}\left(M\right) \, \& \, pOH=2$

$or \, \, \, \left[N a O H\right]=\left(10\right)^{- 2} \, \left(M\right)$

$4 \, ml \, of \, \left(10\right)^{- 2} \, \left(M\right) \, HCl\equiv 4\times \left(10\right)^{- 5} \, moles \, HCl.$

$6 \, ml \, of \, \left(10\right)^{- 2} \, \left(M\right)NaOH\equiv 6\times \left(10\right)^{- 5} \, moles \, NaOH$

After mixing total moles of $\text{OH}^{-}$ $=$ moles of $\text{NaOH} -$ moles of $\text{HCl}$

$= 6 \times 10^{- 5} - 4 \times 10^{- 5}$

$After \, mixing \, excess \, moles \, of \, OH^{-}=2\times 10^{- 5}$

Total volume $= 6 + 4 = 10 \, \text{ml} \, \text{=} \, \text{10} \, \times 10^{- 3} \, \text{litre}$

$\left[O H^{-}\right]=\frac{2 \times 10^{- 5}}{10}\times 10^{3}=2\times 10^{- 3}$

$or \, pOH=3-log 2 = 3 - 0.3 = 2.7$

$or \, \, \, pH=11.3$