Question

 

1) A volume of 80.0 mL of a 0.560 M HNO3 solution is titrated with 0.230 M KOH. Calculate the volume of KOH required to reach the equivalence point.

2) 100. mL of 0.200 M HCl is titrated with 0.250 M NaOH. What is the pH of the solution after 50.0 mL of base has been added? AND What is the pH of the solution at the equivalence point?

3) Determine the pH at the equivalence point for the titration of 40.0 mL of 0.1 M HNO2 by 0.200 M KOH (Example 16.7). (The pKa of nitrous acid is 3.36.)

Answer 1

1) We know, at equivalence point of an acid base titration,

No. of moles acid = No. of moles of base

Now, 0.580 M HNO₃= 0.580 moles in 1000 mL of solution

similarly, 0.230 M KOH= 0.230 moles in 1000 mL of solution

so, in 1000 mL of solution the no of moles of HNO₃=0.580 moles

in 80 mL of solution the no of moles of HNO₃= (0.580*80)/1000 moles

let in "V" mL of 0.230 M KOH is needed to reach the equivalence point.

thus, in 1000 mL of solution no of moles KOH = 0.230 moles

in "V" mL of solution no of moles KOH = (0.230*V)/1000 moles

so in equivalence point,

(0.230*V)/1000 = (0.580*80)/1000

0r, V=(0.580*80)/0.230 mL

=201.74mL

so, 201.74 mL of 0.230 M KOH is needed to reach the equivalence point in course of titration of 80mL of 0.580 M HNO₃

2) No. of moles in 100 mL of 0.200 M HCl is =(100*0.200)/1000 = .02 moles

No. of moles in 50 mL of .250 M NaOH is =( 50*0.250)/1000 = 0.0125 moles

so, after addding 50 mL of 0.250 M of NaOH in 100 mL 0f 0.200 M the no moles unreacted acid in (100+50)=150mL solution is(.02 - 0.0125) moles = 7.5*10^-03 moles

so in 150 mL the no of mole of HCl=7.5*10^-03 moles

in 1000 mL the no of mole HCl =(7.5*10^-03 * 1000)/150 moles

=0.05 moles

i.e the now the strenth of the acid is .05 M

now pH of the solution is = where concentraion of the acid

= - log(.05)

= 1.30

clearly, at equivalence point the acid has been neutralised by the base thus the pH is 7.

3)HNO₂ in water is a weak acid that can be shown by the following equilibrium

HNO₂ + H₂O NO₂^- (aq) + H⁺

now this weak acid is treated by a strong base. Now its pH can be determined by the Henderson's equation which is in following,

where concentraion of the salt and = concentration of acid

now at equilibrium point the acid has been neutralised thus

hence, pH= pKa

= 3.36