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Question: Answered & Verified by Expert
50 mL of 0.2M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be:
ChemistryIonic EquilibriumJEE MainJEE Main 2017 (09 Apr Online)
Options:
  • A 8.25
  • B 9.25
  • C 3.75
  • D 4.75
Solution:
2858 Upvotes Verified Answer
The correct answer is: 9.25

                                          NH3          +                      HCl                           NH4Clmolesat ​​​​​​​ t=0         50×.2×103            25×.2×103           molesat ​​​t=t         25×.2×103                       0                           25×.2×103

 salt & base present in soluton, so it is basic Buffer solution.

pOH=pkb  NH3 +logsaltbase=4.75 pOH=Pkb=4.75

pH=14-pOH=14-4.75=9.25.

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