If 200 mL of a 0.031 molar solution of H2SO4 are added to 84 rall. of a 0.150 M KOH solution, what is the pH of the resulting solution ?
A
12.4
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B
1.7
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C
2.2
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D
10.9
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Solution
The correct option is D 10.9 mmol of H+ (initial) =200×0.031×2=124 mmol of OH− (initial) =84×0.15=12.6 mmol of OH− (left) after neutralisation =0.2[OH−]final=0.2284=7×10−4M pOH=3.15 and pH=10.85