If you have given the ionisation constant of a weak acid and its original concentration, the Ph of the the solution can be calculated.The approach used is the inverse of that followed in the tutorial pH to pKa conversion, here Ka is known and Ph must be calculated.

First, lets see an example to understand this:

Nicotinic acid HC6H4O2N (Ka=1.4 x 10^-5), is another name for niacin , an important member of vitamin group.Determine pH of solution prepared by dissolving 0.10 mol of nicotinic acid,in water to form one liter of solution.

stratefy: To start with, set up an equilibrium table .To accomplish this note that-

the original concentrations of HNic,H+ & Nic- are 0.10 M,0.00M nad 0.00M, respectively ignoring H+ ions from the ionisation of water.

the changes in concentration are related by the coeffcients of the balanced equation, all of which are 1;

Letting delta[H+]= x , it follows that delta[Nic-]=x;delta[HNic]=-x.This information should enable you to express the equilibrium concentration of all species in terms of x.The rest is algebra; substitute into the Ka expression and solve for x= [H+]

solution; setting up the table

HNic (aq) <---------> H+ (aq) + Nic- (aq)

I 0.10 0.0 0.0

C -X +X +X

E 0.10-X X X

substituting into the expression for Ka

Ka = x . x / 0.10-x = 1.4 x 10^-5

this is a quadratic equation .It could be rearranged to form the ax+^2 + bx + c = 0 and solved for x using the qudratic formula.Such a procedure is the time-consumiong and in this case unnecessary.Nicotinic aicd is a weak acid, only slightly ionized in water.The equilibrium concenration of HNic, 0.10-x, is probably only very slightly less than its original concentration.0.10M.So, lets make the approximation 0.10- x = 0.10.This simplifies the equation written above;

x^2 / 0.10 = 1.4 x 10^5

x^2 = 1.4 x 10^-6

taking square root,

x = 1.2 x 10^-3 M = [H+]

pH = -log[H+]

= - log 1.2 x 10^-3

= 2.93