SATHEE: Equilibrium Question 401

Equilibrium Question 401

Question: pH of a solution of $10, m l$ . 1N sodium acetate and $50, m l$ 2N acetic acid $(K_{a} = 1.8 \times 10^{- 5}),$ is approximately [MP PMT 2003]

Options:

A) 4

B) 5

C) 6

D) 7

Show Answer

Answer:

Correct Answer: A

Solution:

$p H = p K_{a} + l o g \frac{[Salt]}{[Acid]}$

$p H = - \log, (1.8 \times 10^{- 5}) + \log \frac{[10]}{[100]}$

$= - \log 1.8 + 5 + \log 10^{- 1}$

$= - 0.2553 + 5 - 1 = 3.7447, o r = 4$

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Equilibrium Question 401

Question: pH of a solution of 10,ml . 1N sodium acetate and 50,ml 2N acetic acid (Ka=1.8×10−5), is approximately [MP PMT 2003]

Options:

Answer:

Solution:

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Question: pH of a solution of $10, m l$ . 1N sodium acetate and $50, m l$ 2N acetic acid $(K_{a} = 1.8 \times 10^{- 5}),$ is approximately [MP PMT 2003]