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 ‌
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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 =965s 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 .