SATHEE: Equilibrium Question 894

Equilibrium Question 894

Question: In which case change in pH is maximum-

Options:

A) 1 mL of pH = 2 is diluted to 100 mL

B) 0.01 mol of $N a O H$ is added to 100 mL of 0.01 M $N a O H$ solution

C) 100 mL of $H_{2} O$ is added to 900 mL of $10^{- 6}$ M $H C l$

D) 100 mL of pH = 2 solution is mixed with 100 mL of pH = 12

Show Answer

Answer:

Correct Answer: D

Solution:

$p H$ before $p H$ after Changes [a] 2 4 2 [b] 12 13.04 1.04 [c] 6 6.7 0.7 [d] 2 7 5 Explanation: [a] $p H = 2. [H^{+}] = 10^{- 2} M$

$M_{1} V_{1} = M_{2} V_{2}$

$10^{- 2} \times 1 = M_{2} \times 100$

$M_{2} = 10^{- 4}$

$∴$ $p H = 4$

[b] $[O H^{-}] = 10^{- 2}$

$p O H = 2$

$p H = 12$ 100 of 0.01 M $N a O H$

contains 0.001 mol $N a O H$ added - 0.01 mol Total moles = 0.011 mol

$∴$ ${[O H^{- 1}]}_{f i n a l} = 0.011 \times 10 = 0.11 M$

$p O H = 0.96$

pH = 13.04 [c] $[H C l] = 10^{- 6} M$

$[H^{+}] = 10^{- 6} M; p H = 6$

After dilution $[H C l] = 10^{- 7} M = [H^{+}]$

$[H^{+}] i n, H_{2} O = 10^{- 7} M$

Total $[H^{+}] = 10^{- 7} + 10^{- 7} = 2 \times 10^{- 7}$

$∴$ $p H = 7 - \log 2 = 6.7$

[d] $, p H = 2,, [H^{+}] = 10^{- 2} M$

$p H = 12,, [H^{+}] = 10^{- 12} M$

$∴$ $, [O H^{-}] = 10^{- 2} M$

Equilibrium Question 894

Question: In which case change in pH is maximum-

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

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Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Equilibrium Question 894

Question: In which case change in pH is maximum-

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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