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?

पर्याय

• iv

• i

• ii

• iii

Answer 1

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