SATHEE: Equilibrium Question 896

Equilibrium Question 896

Question: If acetic acid is 1.0% ionised in its decinormal solution, then pH of its seminormal solution will be

Options:

A) 3.4

B) 4.8

C) 1.6

D) 2.6

Show Answer

Answer:

Correct Answer: D

Solution:

$C H_{3} C O O H, C H_{3} C O O^{-} +, H^{+}$

$\begin{matrix} I n i t i a l, c o n c e ., \frac{1}{10} & 0 & 0 \\ A f t e r, t, t i m e \frac{1}{10} - \frac{1}{10} \times \frac{1.0}{100} & \frac{1}{10} \times \frac{1.0}{100} & \frac{1}{10} \times \frac{1.0}{100} \\ , = 0.1 & , = 0.001 & = 0.001 \end{matrix}$

$K = \frac{0.001 \times 0.001}{0.1}$

$= 1 \times 10^{- 5}$ (i) In case of seminormal solution $C H_{3} C O O H, C H_{3} C O O^{-}, + H^{+}$

$\begin{matrix} I n i t i a l, c o n c . \frac{1}{2} & 0 & 0 \\ A f t e r, t, t i m e \frac{1}{2} - \frac{1}{2} \times \frac{x}{100} & , \frac{1}{2} \times \frac{x}{100}, & \frac{1}{2} \times \frac{x}{100} \end{matrix}$

$= 0.5$ [let solution is $x$ ionised] $K = \frac{{(\frac{1}{2} \times \frac{x}{100})}^{2}}{0.5}$

$= \frac{x^{2}}{20000}$ (ii) From Eqs (i) and (ii) $\frac{x^{2}}{20000} = 1 \times 10^{- 5}$

$x = 0.447$

$[H^{+}] = \frac{1}{2} \times \frac{x}{100}$

$= \frac{1}{2} \times \frac{0.447}{100} = 0.00224$

$p H = - \log, [H^{+}]$

$= - \log, [0.00224] = 2.65$

Equilibrium Question 896

Question: If acetic acid is 1.0% ionised in its decinormal solution, then pH of its seminormal solution will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Equilibrium Question 896

Question: If acetic acid is 1.0% ionised in its decinormal solution, then pH of its seminormal solution will be

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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