Question

The pH of 0.5 L of 1.0 M NaCl after the electrolysis for 965 s using 5.0 A current, is ......?

Answer 1

Equation representing the electrolysis of aqueous NaCl:

$2NaCl(aq) +2H_{2}O(l) --->2 Na^{+}(aq)+2OH^{-}(aq)+H_{2}(g)+ Cl_{2}(g)$

Current I = 5.0 A

Time t = 965 s

Charge Q = It

= 5.0 A (965 s)

= 4,825 C

Moles of NaCl = 0.5 L ×$\frac{1.0 mol}{L}$ = 0.5 mol NaCl

4825 C × $\frac{1 mol e^{-}}{96485C}$×$\frac{2 molNaOH}{2 mol e^{-}}$ = 0.05 mol NaOH

$\frac{0.05 mol}{0.5 L} = 0.1M$

$pOH =-log[OH^{-}] \\ = 1$

pH + pOH = 14.00

pH = 14.00 - 1 = 13

Answer 2

Answer:  pH of solution will be 13.

Explanation: $Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 5A

t= time in seconds = 965 sec

$Q=5A\times 965s=4825C$

$2NaCl+2H_2O\rightarrow 2Na+Cl_2+2OH^-+H_2$

$96500\times 2=193000Coloumb$ of electricity electrolyzes 1 mole of NaCl

4825C of electricity deposits =$\frac{1}{96500}\times 4825=0.05moles$ of NaCl

0.05 moles of naCl will produce 0.05 moles of $OH^-$

Molarity of $OH^-=\frac{\text {no of moles}}{\text {Volume of solution in L}}$

$Molarity=\frac{0.05moles}{0.5L}=0.1M$

$pOH=-log[OH^-]=-log[0.1]=1$

pH+ pOH= 14

$pH=14-1=13$