Question

3. A 40.00 mL 0.100 M solution of HCN was slowly titrated with 0.200 M NaOH solution. Complete the following table and show your calculations in the space provided. Volume of NaOH added (mL) pH 0.00 5.00 10.00 15.00 20.00 25.00 Show your calculations for pH clearly and in steps.

Answer 1

a) M1 = molarity of HCN = 0.1 M

V1 = volume of HCN = 40 ml

M2 = molarity of NaOH = 0.2M

V2 = Volume of NaOH = 0ml

M = Molarity of solution = $\frac{M1V1 + M2V2}{V1+V2}$ pH = -log[H⁺] = -log[total molarity(M)]

M = 0.1 M

pH = 1

b)

M1 = molarity of HCN = 0.1 M

V1 = volume of HCN = 40 ml

M2 = molarity of NaOH = 0.2M

V2 = Volume of NaOH = 5ml

M = Molarity of solution = $\frac{M1V1 + M2V2}{V1+V2}$

M = 0.11111 M

pH = 0.954

c)

M1 = molarity of HCN = 0.1 M

V1 = volume of HCN = 40 ml

M2 = molarity of NaOH = 0.2M

V2 = Volume of NaOH = 10ml

M = Molarity of solution = $\frac{M1V1 + M2V2}{V1+V2}$

M = 0.12 M

pH = 0.921

d)

M1 = molarity of HCN = 0.1 M

V1 = volume of HCN = 40 ml

M2 = molarity of NaOH = 0.2M

V2 = Volume of NaOH = 15ml

M = Molarity of solution = $\frac{M1V1 + M2V2}{V1+V2}$

M = 0.127 M

pH = 0.895

e)

M1 = molarity of HCN = 0.1 M

V1 = volume of HCN = 40 ml

M2 = molarity of NaOH = 0.2M

V2 = Volume of NaOH = 20ml

M = Molarity of solution = $\frac{M1V1 + M2V2}{V1+V2}$

M = 0.1333 M

pH = 0.8751

f) M1 = molarity of HCN = 0.1 M

V1 = volume of HCN = 40 ml

M2 = molarity of NaOH = 0.2M

V2 = Volume of NaOH = 25ml

M = Molarity of solution = $\frac{M1V1 + M2V2}{V1+V2}$

M = 0.1385 M

pH = 0.8587