Chemistry (University)

Question

You have been asked to prepare 1 litre solution of nicotinic acid, HC6H4NO2 (HN), using 0.1 mol of the acid. Determine the pH of this solution (Ka = 1.4x10-5).

(This question is the most important one, plaease help me to answer)

Answer 1

No. of mole= vol*concentration
0.1 =1*X
concen=0.1mol/dm-3
Ka=[H+][H+]/[C6H5NO2]
[H]=Squreroot of:1.4*10-5*0.1
=7*10-6 moldm-3
pH=-log10[H]
=-log7*10-6
=5.15490196

Answer 2

From the other answer, you see that the answer is 7 X 10^ -6.

How about consider purely H2O alone. It is 1 X 10^ -7. In this case, you cannot neglect the ionization of water !!!

HN === H+ + N-
0.1-x *** x+y * x

H2O === H+ + OH-
********** x+y *** y

*** = 空白

(x+y)(x) / (0.1-x) = 1.4 X 10-5 --------------- (1)

(x+y)(y) = 1 X 10-14 ---------------------------- (2)

Solve it and you will get the answer.