Question

A 25.00 mL of 0.0500 M imidazole (B) solution was titrated with 0.1250 M HNO3. The pKa of the imidazolium chloride (BH+) is 6.993.

(a) what is the pH of the solution before adding any HNO3?

(b) What is the pH of the solution after adding 5.00 mL of HNO3?

(c) What is the pH of the solution after adding 10.00 mL of HNO3?

(d) What is the pH of the solution after adding 12.00 mL of HNO3?

Answer 1

Answer: As per the given informations in questions

a] Before addition of HNO3 The Imidazole is present only povide a basic condition

hence we have 0.0500 M imidazole in 1000 ml

hence number of moles in 25 ml = 0.05 * 25 / 1000 = 0.00125 mol

Nad the concentration of Oh- = 0.00125 / 25 * 1000 = 0.05

Hecne POH = - Log [ 0.05] = 1.30103

and ph = 14- 1.30103 = 12.69

Hence the Ph = 12.69

b] The number of moles of 5 ml HNO3 = 0.125 * 5 / 1000 = 0.000625 mol

and moles of base = 0.00125 mol

hence after the reaction the concentration of salt = 0.000625

and the concentration of imidazale = 0.000625

now using hendersons equation we get

POh = Pkb + log [salt]/ [base] = 6.993 + 0 = 6.993

and hence the Ph = 14 - 6.993 = 7.007 = 7

c] Now number of Moles of 10.00 ML HNO3 = 0.125 * 10 / 1000 = 0.00125 mol

moles of base = 0.00125 mol

And hence the complete newutralysation is take place so the Ph = Pka = 6.993

So , the Ph is comes out to be = 6.993

d] number of moles of 12 ml HNO3 = 0.125 * 12 / 1000 = 0.0015

and number of moles of base = 0.00125 mol

Hence after the reaction the remaining number of moles of acid = 0.00025 mol

moles of salt = 0.00125 mol

Ph = Pka + log [salt] / [acid] = 6.993 + log 0.00125 / 0.00025

= 7.6919

hecen the required Ph is comes out to be 7.7 around