BITSAT 2024 Shift 1 Paper

The pH of a 0.5L,1.0M‌NaCl solution after electrolysis for 965 seconds using a 5.0 A current can be determined through the following steps:
Reaction Involved:
NaCl+H2O

‌ Electrolysis ‌

‌───────▶‌

‌

1

2

H2+Cl2+NaOH
Amount of NaCl Present:
In a 0.5 L solution of 1.0 M NaCl , the number of moles of NaCl is 0.5 mol .
Quantity of Electricity Passed:
Current =5.0A
Time =965 s
Total charge passed =965×5 coulombs =4825 coulombs .
Decomposition of NaCl :
Using Faraday's laws, the moles of NaCl decomposed are calculated as:
‌

4825c

96500C∕mol

=0.05‌mol
Formation of NaOH :
The number of moles of NaOH formed will also be 0.05 mol .
Volume of Solution:
Volume =0.5L
Molarity of NaOH :
Molarity of NaOH=‌

0.05‌ moles ‌

0.5L

=0.1M=10−1M
pOH Calculation:
pOH=1
pH Calculation:
pH=14−pOH=14−1=13
Thus, after electrolysis, the pH of the solution will be 13 .