NEET Chemistry · Equilibrium

pH, common ion effect and hydrolysis of salts

Taught in NCERT Class 11 Chemistry, Chapter 6: Equilibrium.

2 verified questions: Easy 0 · Medium 1 · NEET level 1

Sample questions

Sample 1Medium

What is the pH of a 1.0 × 10^-3 M solution of Ba(OH)2 at 298 K, assuming complete dissociation? (log 2 = 0.30, K_w = 1.0 × 10^-14)

A
11.00
B
3.00
C
2.70
D
11.30

Show the answer

The answer is D.

Each Ba(OH)2 gives two OH^- ions, so [OH^-] = 2.0 × 10^-3 M and pOH = 3 − 0.30 = 2.70. pH = 14 − 2.70 = 11.30.

A.
This takes [OH^-] as 1.0 × 10^-3 M, ignoring that Ba(OH)2 gives two OH^- ions.
B.
pH 3.00 is acidic; a base solution cannot have pH below 7.
C.
2.70 is the pOH, not the pH.

NCERT: Class 11 Chemistry, Chapter 6, 6.11

Sample 2NEET level

The pH of a 0.005 M aqueous solution of Ba(OH)2 at 298 K, assuming complete ionisation (Kw = 1.0 × 10⁻¹⁴), is

A
2.0
B
11.7
C
12.0
D
2.3

Show the answer

The answer is C.

Each Ba(OH)2 gives two OH⁻ ions, so [OH⁻] = 2 × 0.005 M = 0.01 M and pOH = 2. Since pH + pOH = 14 at 298 K, pH = 14 − 2 = 12.0.

A.
This is the pOH, not the pH.
B.
Takes [OH⁻] = 0.005 M, forgetting that each Ba(OH)2 gives two OH⁻ ions.
D.
This is the pOH computed with [OH⁻] = 0.005 M; it is neither pH nor correct.

NCERT: Class 11 Chemistry, Unit 6 (Equilibrium), The pH Scale

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pH, common ion effect and hydrolysis of salts

Sample questions

Practise pH, common ion effect and hydrolysis of salts

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