Question

Consider the titration of 35.0 mL of 0.260 M HF with 0.215 MNaOH. Calculate the pH at each of the following points.

Part A.)

How many milliliters of base are required to reach the equivalence point?

Express your answer using three significant figures.

Part B.)

Calculate the pH after the addition of 8.75 mL of base

Express your answer using two decimal places.

Part C.)

Calculate the pH at halfway to the equivalence point.

Express your answer using two decimal places.

Part D.)

Calculate the pH at the equivalence point.

Express your answer using two decimal places.

Part E.)

Calculate the pH after the addition of 70.0 mL of base.

Express your answer using two decimal places.

Answer #1

HF ,

Ka = 6.3 x 10^-4

pKa = 3.20

part A )

millimoles of acid = millimoles of base

35 x 0.260 = V x 0.215

V = 42.3 mL

**volume of base = 42.3 mL**

part B)

millimoles of acid = 35 x 0.260 = 9.1

millimoles of base = 0.215 x 8.75 = 1.88

HF + NaOH ----------------> NaF + H2O

9.1 1.88 0 0

7.22 0 1.88 1.88

pH = pKa + log [NaF]/[HF]

pH = 3.20 + log (1.88 / 7.22)

**pH = 2.62**

part C)

half equivalecnce point : pH = pKa

**pH = 3.20**

part D )

equivalece point:

salt only formed here

salt concentration = 9.1 / 35 + 42.3 = 6.15 x 10^-3 M

pH = 7 + 1/2 [pKa + logC] = 7 + 1/2 (3.20 + log (6.15 x 10^-3))

**pH = 7.49**

part E )

millimoles of base = 70 x 0.215 = 15.05

base remaining = 15.05 - 9.1 = 5.95

base concentration = 5.95 / (35 +70) = 0.057 M

pOH = -log [OH-] = -log(0.057)

pOH = 1.25

pH + pOH =14'

**pH = 12.75**

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