Question

A solution is made by combining 50 mL 1.0 M carbonic acid, 2.0 mL 5.0 M KOH and 448 mL pure water (assume the total volume is 500 mL). The pKa of carbonic acid is 6.35. Calculate the pH of the resulting solution.

Answer 1

No. of mmole H2CO3 = 50 mL x 1.0 mmol/mL = 5.0 mmol H2CO3
No. Of mmol KOH = 2.o mL x 5.0 mmol KOH/mL = 10. mmol KOH

equation: H2CO3 + KOH -------> KHCO3 + H2O
start...........50mmol..10mmol............0..............0

finally.........40. mmol...0 mmol...........10. mmol......10 mmol

[KOH] is limiting reagent as it is less than [H2CO3]

pH = pKa + log10 [Cs]/[Ca] : Cs = molarity HCO3-. Ca = molarity H2CO3
Since the volume is the same we can use mmol instead of molarity

pH = pKa + log ( [salt]/[acid])

pH = 6.36 + log (10 mmol HCO3-/40 mmol H2CO3) = 5.76