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Question
50 ml of 0.05 M HNO3 is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.
Solution
Number of moles of HNO3 = 0.05 × 50 × 10−3
= 2.5 × 10−3
Number of moles of KOH = 0.025 × 50 × 10−3
= 1.25 × 10−3
Number of moles of HNO3 after mixing
= 2.5 × 10−3 – 1.25 × 10−3
= 1.25 × 10−3
∴ Concentration of HNO3 = `("Number of moles of HNO"_3)/("Volume is litre")`
After mixing, total volume = 100 ml = 100 × 10−3 L
∴ [H+] = `(1.25 xx 10^-3 "moles")/(100 xx 10^-3 "L")`
= 1.25 × 10−2 moles L−1
pH = –log [H+]
pH = –log (1.25 × 10−2)
= 2 – 0.0969
= 1.9031
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