Question

What is the pH at the equivalence point in the titration of a 23.0 mL sample of a 0.304 M aqueous hydrocyanic acid solution with a 0.433 M aqueous sodium hydroxide solution?

pH =

A 30.6 mL sample of a 0.393 M aqueous acetic acid solution is titrated with a 0.296 M aqueous potassium hydroxide solution. What is the pH after 16.7 mL of base have been added?

pH =

When a 18.0 mL sample of a 0.476 M aqueous acetic acid solution is titrated with a 0.320 M aqueous potassium hydroxide solution, what is the pH after 40.2 mL of potassium hydroxide have been added?

pH =

Answer 1

a)

Hydrocyanic acid = HCN = 23.0 mL of 0.304M

number of moles of HCN = 0.304M x 0.023L = 0.00699 moles

Concentration of NaOH = 0.433M

at equivalent point

volume of NaOH = 0.304 M x 23.0mL / 0.433 = 16.15 mL [ M1V1 = M2V2 ,          V2 = M1V1/M2]

at equivalent point

number of moles of acid is equal to base

number of moles = 0.00699 moles

total volume = 23.0 + 16.15 = 39.15 mL = 0.03915 L

Ka of HCN = 6.2 x10^-10

Ka = 6.2 x10^-10

-log( Ka) = -log(6.2 x10^-10)

PKa = 9.21

at equivalent point

PH = 7 + 1/2[PKa + LogC]

where C= number of moles / volume in L = 0.00699 / 0.03915 = 0.178M

PH = 7 + 1/2 [ 9.21 + log( 0.178)]

PH = 11.23,

b)

acetic acid solution = 30.6 mL of 0.393M

number of moles of acetic acid = 0.393M x 0.0306L = 0.0120 moles

Potassium hydroxide = 16.7 mL of 0.296M

number of moles of KOH = 0.296M x 0.0167L = 0.00494 moles

Ka of acetic acid = 1.8 x10^-5

Ka = 1.8 x10^-5

-log( Ka) = -log( 1.8 x10^-5)

PKa = 4.74

                              CH3COOH       +        KOH --------------------- CH3COOK    + H2O

Initial                0.0120                          0.00494                             0

change          - 0.00494                        - 0.00494                      + 0.00494

remaining        0.00706                               0                             + 0.00494

PH = Pka + log[salt]/[acid]

PH = 4.74 + log( 0.00494/ 0.00706)

PH = 4.58

c)

acetic acid solution = CH3COOH = 18.0mL of 0.476M

number of moles of CH3COOH = 0.476 M x 0.0180L = 0.008568 moles

KOH = 40.2mL of 0.320M

number of moles of KOH = 0.320M x 0.0402L = 0.01286moles

number of moles of KOH is greaterthan acid

so, the nature of the solution is basic

remaing number of moles of KOH = 0.01286 - 0.008568 = 0.004292 moles

number of moles of OH- = 0.004292 moles

total volume = 18.0 mL + 40.2 mL = 58.2mL = 0.0582 L

Concentration of OH- = number of moles / volume in L = 0.004292 / 0.0582 = 0.0737 M

[OH-] = 0.0737m

-log[OH-] = -log( 0.0737)

POH = 1.13

PH + POH = 14

PH = 14 - POH

PH = 14 - 1.13

PH = 12.87