Answer to Question #254005 in General Chemistry for faida

Question

calculate the ph of a solution prepared by mixing 80 ml of 0.2 m kh2po4 and 120 ml of 0.2 m k2hpo4

Answer 1

Using HH equation

pH = pKa + log (Base/Acid)

(Base) KH2PO4 = (0.2 mmol/mL) x (120 mL) = 24 mmol

(Acid) K2HPO4 = (0.2 mmol/mL) x (80 mL) = 16 mmol

You can use mmol directly because the units cancel.

I got the pKa value from Buffers

pH = 7.21 + log(24/16) = 7.21 + 0.176 = 7.386 = 7.4