In which case change in pH is maximum ?
A) 1 mL of pH = 2 is diluted to 100 mL
B) 0.01 mol of NaOH is added into 100 mL of 0.01 M NaOH solution
C) 100 mL of H,O is added into 900 mL of 10^-6 M HCI
D) 100 mL of pH = 2 solution is mixed with 100 mL of pH = 12
answer is option D.. But I want clear explanation
Answers
Answer:
(i) pOH=−log[OH
−
]
(ii) pH+pOH=14
∵100mL0.01MNaOH solution is diluted to 1dm
3
(ie, 10 times diluted)
∴ Resultant solution of NaOH=0.001M
NaOH⟶Na
+
+OH
−
∴[OH
−
]=0.001=10
−3
pOH=−log[OH
−
]
=−log10
−3
=3
pH+pOH=14
∴pH=14−pOH=14−3=11
Answer:
Option D- 100 ml of pH=2 solution is mixed with 100 ml of pH=12
is correct.
Explanation:
Option A: 1ml of pH=2 is diluted to 100 ml.
pH=2
[H⁺]=10⁻²
Volume= 100ml
We find the pH after diluting using,
M₁V₁=M₂V₂
10⁻²×1=M₂×100
M₂=10⁻⁴
pH=4
Option B- 0.01 mol of NaOH is added into 100 ml of 0.01 M NaOH solution.
[OH⁻]=10⁻²
pOH=2
pH=12
In 100 ml of 0.01 M NaOH solution, we have 0.001 mol and when 0.01 mol of NaOH is added the total number of moles will be 0.011mol.
Then,
[OH⁻]= 0.011×10=0.11 M
pOH= 0.96
Final pH= 14-0.96
= 13.04
Option C- 100ml of H₂O is added into 900ml of 10⁻⁶ M HCl.
[H⁺]= 10⁻⁶ M
Final pH= 6
After the addition of water, the molarity of HCl will be 10⁻⁷M which is the molarity of [H⁺] too.
The molarity of [H⁺] in H₂O is 10⁻⁷M.
Total [H⁺]=10⁻⁷+10⁷
= 2 × 10⁻⁷
Final pH= 6.7
Option D- 100 ml of pH=2 solution is mixed with 100 ml of pH=12.
pH= 2
[H⁺]=10⁻²
pH= 12
[H⁺]=10⁻¹²
The [OH⁻]= 10⁻²
Final pH= 7
Conclusion:
The difference between the initial pH and final pH is largest in D.
The correct option is D- 100ml of pH=2 solution is mixed with 100 ml of pH=12.