Question

50 mL of 2.0 M NaOH is added to a 1.0L buffer solution that is 0.20M in HF and 0.20M in KF. The ka for HF is 3.5 x 10^-4.

1. Write the reaction or the buffer only including states and excluding spectator ions.

2. Write the neutralization reaction that occurs after the addition of LiOH including states and excluding spectator ions.

3. Calculate the amount of moles remaining and the final concentration for each component of the solution after the neutralization reaction occurs.

4. Calculate the resulting pH of the buffer - be sure the include validation if approximation is used.

Answer 1

1) HF(aq) + H2O(l) <------> H3O+(aq) + F-(aq)

Ka = [ H3O+ ] [ F- ] / [ HF ] = 3.5×10^-4

2) added OH- ion react with HF

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

3) No of mole of OH- ion = (2mol/1000ml)×50ml = 0.1mole

0.1 mol of OH- react with 0.1mole of HF

No of mol of HF before addn=0.2mol

No of mol of HF remaining = 0.2mol - 0.1 mol =0.1mol

No of mole of F- before addn = 0.2mol

No of mole of F- after addn = 0.2mol + 0.1mol = 0.3mol

Total volume = 1000ml + 50ml = 1050ml

[ HF ] = (0.1/1050ml)×1000ml = 0.0952M

[ F- ] = ( 0.3/1050ml)× 1000ml = 0.2857M

4) Henderson equation is

pH = pKa + log ( [ A- ] / [ HA ] )

Concentration of Acid, [ HA] = [ HF] = 0.0952M

Concentration of Conjugate base , [ A- ] = [ F- ] = 0.2857M

Ka = 3.5×10^-4

pKa = - log ( Ka)

pKa = 3.46

Therefore,

pH = 3.46 + log ( 0.0952/0.2857)

= 3.46+0.48

= 3.94