Question

Consider the titration of 39.2 mL of 0.260 M HF with 0.215 M NaOH. Calculate the pH at each of the following points.

A. Calculate the pH after the addition of 9.80 mL of base

B.Calculate the pH at halfway to the equivalence point.

C. Calculate the pH at the equivalence point.

Answer 1

Calculate moles of HF

n HF = volume in L x molarity = 0.0392 L x 0.260 M =0.010192 mol

calculate moles of base NaOH

n NaOH = 0.00980 L x 0.215 M = 0.002107 mol

write reaction between HF and NaOH

NaOH (aq) + HF (aq) ---- > NaF (aq) + H2O (l)

Mol ratio between NaOH and HF is 1:1

Calculate moles of NaOH needed to react with 0.01092 mol HF

n NaOH = 0.01092 mol HF x 1 mol NaOH / 1 mol HF

= 0.01092 mol NaOH
moles of NaOH are 0.002107 so it is limiting reactant.

And therefore NaF formed = number of moles of NaOH (mol ratio between NaOH and NaF is 1:1)

Lets calculate remaining moles of HF

n mol HF = initial moles of HF – moles of HF reacted with NaOH

=0.010192 – 0.002107 mol = 0.008086 mol acid

Calculate equilibrium concentration of acid and NaF

[HF] = 0.008086 mol / total volume in L = 0.008086 mol / (0.0392+0.00982) = 0.1649 M

[NaF]= 0.02107 / (0.0392+0.00982 ) = 0.4298 M

We use Henderson Hasselback equation

pH = pka + log ( [NaF]/[HF])

pka of HF = 3.17

pH = 3.17 + log ( 0.4298 / 0.1649)

= 3.59

pH of this solution would be 3.59

B.

At halfway, [NaF]=[HF]

Second term in Henderson Hasselbach equation.

So log ([NaF]/[HF]) = log 1 = 0

pH = pka + log 1

pH = 3.17 + 0 = 3.17

C. At equivalence point, moles of acid = moles of base.

So total number of acid is reacted with base completely.

Moles of NaF formed = moles of acid = 0.010192 mol.

Calculate the molarity of NaF

Molarity = 0.010192 mol / (0.0392+0.00982)

= 0.2079 M

Set up ice

[NaF] = [F-]

   F- (aq) + H2O (l) --- > HF (aq) + OH-(aq)

I   0.2079                     0              0

C   -x                           +x             +x

E (0.2079-x)                  x                 x

Use kb expression and find the value of x

Kb = 1.0 E-14 / Ka

Ka of HF = 6.6E-4

Lets plug this value

Kb = 1.0E-14 / 6.6E-4=1.51E-11

Kb = x^2/(0.2079-x)

1.51E-11 = x^2 / (0.2079-x)

Lets solve this quadratic equation.

1.51E-11 x (0.2079-x) = x^2

x = 1.77 x 10^-6

[OH-]= 1.77 E -6

pOH = - log 1.77 E-6

pOH = 5.75

pH = 14-5.76 = 8.25

pH of the solution = 8.25