Question

a current of 0.193amp is passed through 100ml of 0.2m Nacl for an hour.calculate pH of solution after electrolysis if current efficiency is 90%.assume no volume change?(ans=12.82)(electro chemistry)

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Q}=\mathrm{It} \\ \mathrm{or},\mathrm{Q}=0.193\mathrm{A}\times 60\times 60\mathrm{s}=694.8\mathrm{C} \\ \mathrm{Actual}\mathrm{amount}\mathrm{of}\mathrm{charge}\mathrm{required}\mathrm{as}\mathrm{efficiency}\mathrm{is}90\%=\frac{90}{100}\times 694.8=625.32\mathrm{C} \\ \mathrm{Now},2\mathrm{NaCl}+2{\mathrm{H}}_2\mathrm{O}\leftrightarrow 2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)+\hspace{0.17em}{\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Cl}}_2\left(\mathrm{g}\right) \\ \mathrm{Moles}\mathrm{of}\mathrm{NaCl}=\frac{\mathrm{Molarity}\times \mathrm{volume}}{1000}=\frac{0.2\times 100}{1000}=0.02\mathrm{mol} \\ \mathrm{Moles}\mathrm{of}\mathrm{NaCl}\mathrm{reacted}=\frac{2\times 625.32}{2\times 96500}=0.00648\mathrm{mol} \\ \mathrm{Mol}\mathrm{of}{\mathrm{OH}}^{-}\mathrm{produced}=0.00648\mathrm{mol} \\ \mathrm{Molarity}\mathrm{of}{\mathrm{OH}}^{-}\mathrm{ions}=\frac{0.00648\times 1000}{100}=0.0648 \\ \mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log (0.0648)=1.18 \\ \mathrm{pH}=14-\mathrm{pOH}=14-1.18=12.82 \end{gathered}$$