Question

100 ml solution of 0.1 n hcl was titrated with naoh solution. The titration was diluted after adding 30 ml of naoh. What is volume

Answer 1

Hey mate here's your answer

• pH of 0.1 N H2SO4= - log(0.1)= 1
• The concentration of KOH=0.1 M =0.1N, because KOH is a mono acidic base.
• At HALF NEUTRALISATION, ie., When 50 ml.of KOH, is added it would neutralise 50 ml.of 0.1N HCl and the resulting Solution will have a volume of 150 ml, with (50 x 0.1)= 5 m.moles of HCl in it.
• So 5 m.moles of HCl will be present in 150 ml.of the solution.
• So 1000 ml of this solution will contain= 1000 x 5/150= 33.33 m.mole.
• The concentration of the solution at HALF NEUTRALISATION will be 33.33/1000= 0.0333 N.
• pH at this point= - log[0.0333]=1.477. So at HALF NEUTRALISATION there will not be much change in pH.
• However if we add 100.1 ml of NaOH to 100 ml of the acid Solution with the concentration of 0.1N, then 100 ml HCl will be neutralised by 100 ml of NaOH. There will extra 0.1 ml.of 0.1N NaOH in the volume of (200.1 ml =~ )200 .ml.
• The amount of NaOH in 200 ml= 0.1 x 0.1= 0.01 m.mole. Amount of NaOH present in 1000 ml =0.01x5=0.05 m.mole= 0.05/1000=0.00005= 5 x10^-5 mole
• So pOH after just NEUTRALISATION =- log[5 x 10^-5] =4.301.
• So pH=14–4.301=9.699. So just 2(0.1 ml) drops extra after neutralisation increases the pH to 9.699 !!!.