Question

QUESTION 9 Calculate the pH during the titration of 25.00 ml of 0.1000 M HF(aq) with 0.1000 M ROH(aq) after 9 mL of the base have been added. Ka of HF = 7.4 x 10-4

Answer 1

HF + RbOH ---------> RbF + H2O

Initial

Calculation for finding number of moles:

Number of moles = Molarity x Volume in L

Initial number of moles of HF = 0.0250 L x 0.1000M/ 1L = 0.0025 moles = 2.5 mmol

Number of moles of RbOH added = 0.016 L x 0.1000M/ 1L = 0.0016 moles = 1.6 mmol

Moles of HF remaining = 2.5 mmol - 1.6 mmol = 0.9 mmol

We know that, pKa = -log Ka

So, pKa = - log (7.4 x 10-4) = - (-3.13) = 3.13

pH = pKa + log ([RbF]/[HF]) = 3.13 + log (1.6 mmol/0.9 mmol) = 3.13 + log 1.77 = 3.13 + 0.247 = 3.377