92.0k views
2 votes
What is the ph of a solution of 0.450 m kh2po4, potassium dihydrogen phosphate?

User Landry
by
8.4k points

1 Answer

4 votes
pKa = -log (Ka) = log [HPO4(2-)] - log[H+]^2 = - log(4.2×10^-13)
pH = - log [H+]
- log [H+]^2 = - 2 log [H+]
2pH = - log (4.2×10^-13) - log [HPO4(2-)]
2pH = - log (4.2×10^-13) - log (0.550)
pH = 6.32

User Wueb
by
7.9k points

Related questions

2 answers
0 votes
146k views
1 answer
4 votes
152k views
1 answer
2 votes
193k views
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.