Read the following passage carefully and answer the questio…

Chemistry Ionic Equilibrium General Numeric Response

Published by: CGP EDU Academic Team

Published on: August 13, 2026

Read the following passage carefully and answer the questions.

Consider the following experiment set-up with stopcocks-S 1 ,S 2 ,S 3 ,S 4 and taps-T 1 ,T 2 ,T 3 ,T 4 .

(i) In which of the following cases will the solutions in tub have a pH different from others ?

T 1 and T 4 are opened. S 1 S 4 ; Yes

Only T 2 is opened. S 2 No ; S 4

T 1 , T 2 , T 3 and T 4 are all opened. S 3 S 3 ; No ; S 1

T 1 is opened and the content of bucket is poured in the tub (ii) A pH meter is immersed in the bucket of the original set-up shown. It shows a reading of 7. Now, S 1 is first opened and all content is allowed to drain out into the bucket. pH of the solution in the bucket is measured again. Similarly S 2 , S 3 and S 4 are sequentially opened and the the pH of the solution in the bucket is measured each time after complete drainage. Opening which stop cock would cause maximum pH change (increase/deacrease) compared to the previous pH in bucket ? S 4 (iii) The above apparatus was setup to study the functioning of the human stomach, where a low pH is maintained at a nearly constant value even when reactions take place in it. Following the same procedure as in the previous question, can a solution imitating stomach-like pH properties be generated in the bucket, and if so, by opening which stopcock ? Would this solution have the lowest pH possible from this setup ? If your answer is "No" for any of these questions, then after answering "No", choose the stopcock, opening of which will lead to solution in bucket with minimum pH, following the same procedure as above. S 3 ; No ; S 4

Text Solution

Verified by Experts

The correct answer is:

(i) In all the other cases, pH = 1.

When all taps are opened,

Millimoles of H + = 100

Volume of solution = 2000 mL

∴ [H + ] = = 0.05 M.

pH = – log[H + ] = – log (5×10 –2 ) = 2 – log 5 1.

(ii)

pH i = 7.

After opening S 1 , solution in bucket would contain [CH 3 COOH] = 0.1 M. So, pH after opening
S 1 = (pK a – logC) = 2.8775.

After opening S 2 , solution in bucket would contain [CH 3 COONa] = M. So, pH after opening
S 2 = (pK w + pK a + logC) = 8.7875.

After opening S 3 , solution in bucket would contain [CH 3 COONa] = M & [CH 3 COOH] = M. So, pH after opening S 3 = pK a + log = pK a = 4.755. (Buffer properties)