Question

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)

• A. 10.3
• B. 11.3
• C. 11
• D. 4.3

Answer 1

Answer: (B)

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

$or\left[NaOH\right]={\left(10\right)}^{-2}\left(M\right)$

$4mlof{\left(10\right)}^{-2}\left(M\right)HCl\equiv 4\times {\left(10\right)}^{-5}molesHCl.$

$6mlof{\left(10\right)}^{-2}\left(M\right)NaOH\equiv 6\times {\left(10\right)}^{-5}molesNaOH$

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

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

$AftermixingexcessmolesofO{H}^{-}=2\times 10^{-5}$

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

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

$orpOH=3-log2=3-0.3=2.7$

$orpH=11.3$