0

Guest

Courses New
  • Forum
  • AIPMT
  • In a basic buffer 0.0025 mole of NH4Cl and...

In a basic buffer 0.0025 mole of NH4Cl and 0.15 mole of NH4OH are present. The pH of the solution will be[pKa= 4.74] 1) 11.04 2) 10.24 3) 6.62 4) 5.48

In a basic buffer 0.0025 mole of NH4Cl and 0.15 mole of NH4OH are present. The pH of the solution will be[pKa= 4.74] 1) 11.04 2) 10.24 3) 6.62 4) 5.48

Grade:11

1 Answers

Aman Rajput
24 Points
5 years ago
Taking pKb=4.74 [a misprint in the given data]Using formula- p(OH) = pKb + log(salt/base) = 4.74 + log(0.0025/0.15) = 4.74 + log(0.01666) = 4.74 + [-1.78] = 4.74 - 1.78 = 2.96So, p(OH) = 2.96Since, pH + p(OH) = 14pH = 14 - p(OH) = 14 - 2.96 = 11.04So, the pH of the given solution comes out to be 11.04

Think You Can Provide A Better Answer ?

Other Related Questions on AIPMT

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More