What is the ph of a solution of 0.450 m kh2po4, potassium dihydrogen phosphate?
1 answer:
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
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