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Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations. i. 60 mL MM10 HCl + 40 mL MM10 NaOH ii. 55 mL MM10 HCl + 45 mL MM10 NaOH - Chemistry

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प्रश्न

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?

विकल्प

  • iv

  • i

  • ii

  • iii

MCQ
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उत्तर

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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अध्याय 8: Ionic Equilibrium - Evaluation [पृष्ठ २८]

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सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 8 Ionic Equilibrium
Evaluation | Q 2. | पृष्ठ २८
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