Question

What is the [H3O + ] concentration of a solution formed by mixing 50.0 mL of 0.20 M HCl and 50.0 mL of 0.10 M NaOH?

Given 100.0 mL of a buffer that is 0.50 M in HOCl and 0.40 M in NaOCl, what is the pH after 10.0 mL of 1.0 M NaOH has been added? (Ka for HOCl = 3.5  108 ).

What is the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF? (The pKa of HF is 3.14.)

Answer 1

Molarity = Number of moles / Volume (L)

Now, 50.0 mL of 0.20 M HCl

Number of moles of HCl = 0.05 L x 0.2 M = 0.01 mol of HCl

50.0 mL of 0.10 M NaOH

Number of moles of NaOH = 0.05 L x 0.1 M = 0.005 mol of NaOH

0.005 mol of NaOH will neutralize 0.005 mol of HCl.

So, remaining mole of HCl = 0.01 - 0.005 = 0.005 mol of HCl

Total volume = 50 mL + 50 mL = 100 mL = 0.1 L

[HCl] = 0.005 mol / 0.1 L = 0.05 M

HCl is a strong acid and it will dissociate completely.

So [H₃O⁺] = 0.05 M

pH = -log [H₃O⁺] = - log (0.05) = 1.30

(b)

HOCl + H₂O <-> H₃O⁺ + OCl^-

Ka = ([H₃O⁺][OCl^-]) / [HOCl] = 3.5 x 10^-8

So:

(Ka x [HOCl]) / [OCl-] = [H₃O⁺]

The reaction between HOCl and NaOH is

HOCl + NaOH -> H₂O + Na⁺ + OCl^-

We see that for every mole of HOCl that reacts with NaOH, the number of moles of HOCl will decreases and the number of moles of OCl^- increases.

We can work out the initial numbers of moles of HOCl and OCl^- from the equation:

Conc. = (number of moles) / Volume

For HOCl we have

(0.50 M) x (0.1 L) = 0.05 moles HOCl

For OCl^- we have

(0.40 M) x (0.1 L) = 0.04 moles OCl^-

We now calculate the number of moles of NaOH added:

(1.0 M) x (0.01 L) = 0.01 moles NaOH

We know from the reaction that the number of moles of HOCl will decrease after addition of NaOH and that the number of moles of OCl^- will increase after the addition of NaOH. The number of moles will increase or decrease by 0.01 moles (The number of moles of NaOH added)

Now we calculate the number of moles of HOCl and OCl^- after the addition of NaOH

For HOCl

(Initial moles HOCl) - (Moles NaOH added) = Final number of moles of HOCl

= 0.05 - 0.01

= 0.04 moles HOCl

For OCl^-

(Initial moles OCl^-) + (Moles NaOH added= Final number of moles OCl^-

= 0.04 + 0.01

= 0.05 moles OCl^-

To calculate the final concentrations of HOCl and OCl^- we divide the number of moles by the total volume after the addition of NaOH

= 100 mL + 10 mL

=110 mL

= 0.110 L

So

[HOCl] = 0.364 M

[OCl^-] = 0.454 M

Substituting these values in the expression

(Ka x [HOCl]) / [OCl^-] = [H³O⁺]

We can calculate [H₃O⁺]

= (3.5 x 10^-8 x 0.364) / 0.454

= 2.8 x 10^-8 M

pH = -log [H₃O⁺] = -log (2.8 x 10^-8)

= 7.55

(c)

Moles of HF = 0.05 L x 0.1 M = 0.005 mol of HF

Moles of NaF = 0.025 L x 0.2 M = 0.005 M

Total volume = 50 mL + 25 mL = 75 mL = 0.075 L

Final concentration of HF = 0.005 /0.075 = 0.067 M

Final concentration of NaF = 0.005 /0.075 = 0.067 M

Using Henderson-Hasselbalch equation

pH = pKa + log ([Salt] / [Acid])

     = 3.4 + log(1)

     = 3.4 + 0

     = 3.4