Question

Following solutions were prepared by mixing different volumes of NaOH of HCl different concentrations.

i. 60 mL ${\displaystyle \frac{\text{M}}{10}}$ HCl + 40 mL ${\displaystyle \frac{\text{M}}{10}}$ NaOH

ii. 55 mL ${\displaystyle \frac{\text{M}}{10}}$ HCl + 45 mL ${\displaystyle \frac{\text{M}}{10}}$ NaOH

iii. 75 mL ${\displaystyle \frac{\text{M}}{5}}$ HCl + 25 mL ${\displaystyle \frac{\text{M}}{5}}$ NaOH

iv. 100 mL ${\displaystyle \frac{\text{M}}{10}}$ HCl + 100 mL ${\displaystyle \frac{\text{M}}{10}}$ NaOH

pH of which one of them will be equal to 1?

पर्याय

• iv

• i

• ii

• iii

Answer 1

iii

Explanation:

75 mL ${\displaystyle \frac{\text{M}}{5}}$ HCl + 25 mL ${\displaystyle \frac{\text{M}}{5}}$ 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 = ${\displaystyle \frac{\text{No of moles of HCl}}{\text{Vol in litre}}}$

= ${\displaystyle \frac{10\times {10}^{-3}}{100\times {10}^{-3}}}$

= 0.1 M

for (iii) solution, pH of 0.1 M HCl = –log₁₀ (0.1) = 1