Question

3. A solution is prepared by treating 50 mL of 0.50 M HF (Ka = 6.6x10 ) with 10 mL of 1.15 M NaOH. Calculate the pH of the solution. 4. If enough NaOH is added to the solution in problem 3 to neutralize the HF, what will the pH be? 5. 42.16 mL of a solution of NaOH are required to neutralize 0.7878 g of potassium hydrogen phthalate (Mwt = 204.23 g/mol). Calculate the molar concentration of NaOH in the solution. 6. Calculate the pH of a solution prepared from .0421 mol NH2OH(hydroxylamine, Kb = 6.6x10 ") and 0.028 mol NH,OH 7. If 3 mL of 4 M HCl is added to the above solution, what will the new pH be?

Answer 1

3.

NaOH reacts with HF as : NaOH + HF = NaF + H2O

So, when NaOH is added to HF, the salt NaF is formed, and some HF is left over .

The resulting solution thus contains a weak acid (HF) and its salt (NaF). This is an acidic buffer solution. The pH of buffer solutions is calculated using the Henderson-Hasselbalch equation :

[Salt] pH = pka + lograridi

Ka is acid dissociation constant, corresponds to the acid dissociation of : HF → F- + H+

pKa = -log (Ka)= -log(6.6 x 10-4) = 3.18

[…] denotes concentration.

To find concentration of salt and acid:

Number of moles =Molarity x volume added (in L)

So, volume of HF = 50mL = 0.05 L, volume of NaOH = 10mL = 0.01L, total volume = 0.05 + 0.01 = 0.06L

So, moles of HF present = 0.50 M x 0.05 L = 0.025 moles

Moles of NaOH added = 1.15 M x 0.01L = 0.0115 moles

0.0115 moles of NaOH will react with 0.0115 moles of HF to give 0.0115 moles of NaF

Moles of HF left = 0.025 – 0.0115 = 0.0135 moles

Molarity = Number of moles/volume.

So, final molarity of acid (HF) (the total volume of solution is considered now)= 0.0135/0.06L = 0.225 M

Final molarity of salt (NaF) = 0.0115 moles/0.06 L = 0.19167 L

Putting these values in the Henderson equation :

pH = 3.18 + log(0.19167)/(0.225)

Solving, pH = 3.11