Question

A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting solution?

Answer 1

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

          = 0.15 x 0.3/0.8

               = 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

             = 0.10 x 0.5/0.8

                    = 0.0625 M

Note: Total volume = 500 + 300 = 800 mL = 0.8 dm3

pH = 3.14266 + log [0.05625/0.0625]

      = 3.14267 + (-0.04575749056)

           = 3.09691250944

From all the available options below:

a) 2.97

b) 3.10

c) 3.19

d) 3.22

e) 3.32

The correct option is b.