Question

(a) A 25.0mL of a unknwon concentration of HBr solution is titrated with 0.100M NaOH solution. The equivalence point is reached upon the addition of 18.58mL of the base. What is the concentration of HBr solution?

(b) A sample of 10.0mL of 0.200M hydrocyanic acid (HCN) is titrated with 0.299M NaOH. The pKa for hydrocynanic acid is 9.31. what volume of NaOH solution is required to reach the equivalence point of titration?

(c ) Still consider the solution in (b), what is the concentration of NaCN at the equivalence point of titration?

(d) what is the pH of the above solution?

Let's consider titration of 0.1M acetic acid ( lets write its formula as HAc, this is a weak acid with pKa for HAc is 4.74) with 0.1M NaOH.

(e) What is the pH of a 50.0mL sample of 0.1M of acetic acid? Show your calculations.

(f) After stiochiometric equilvalent amount of NaOH is added to the sample (essentially you add another 50.0mL of 0.1M NaOH into it), what compounds do you have in the sample? what is the pH of the solution at the equivalence point of titration?

(typed responses please, hard to interpret most handwriting)

Answer 1

Answer – a) We are given, volume of HBr = 25.0 mL , [NaOH] = 0.100 M

The volume of base required for equivalence point = 18.58 mL

We know the balanced reaction between HBr and NaOH

HBr + NaOH -----> NaBr + H₂O

Now we need to calculate the moles of NaOH

Moles of NaOH = 0.100 M * 0.01858 L

                               = 0.001858 moles of NaOH

From the balanced reaction and we also know at equivalence point moles of acid and base are equal

Moles of NaOH = moles of acid = 0.001858 moles

Molarity of HBr = 0.001858 moles / 0.025 L

                            = 0.0743 M

so, 0.0743 M is the concentration of HBr solution is titrated with 0.100M NaOH solution.