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Question
Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.
i. 60 mL `"M"/10` HCl + 40 mL `"M"/10` NaOH
ii. 55 mL `"M"/10` HCl + 45 mL `"M"/10` NaOH
iii. 75 mL `"M"/5` HCl + 25 mL `"M"/5` NaOH
iv. 100 mL `"M"/10` HCl + 100 mL `"M"/10` NaOH
pH of which one of them will be equal to 1?
Options
iv
i
ii
iii
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Solution
iii
Explanation:
75 mL `"M"/5` HCl + 25 mL `"M"/5` NaOH
No of moles of HCl = 0.2 × 75 × 10−3 = 15 × 10−3
No of moles of NaOH = 0.2 × 25 × 10−3 = 5 × 10−3
No of moles of HCl after mixing = 15 × 10−3 – 5 × 10−3 = 10 × 10−3
∴ Concentration of HCl = `"No of moles of HCl"/"Vol in litre"`
= `(10 xx 10^-3)/(100 xx 10^-3)`
= 0.1 M
for (iii) solution, pH of 0.1 M HCl = –log10 (0.1) = 1
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