how do u calculate pH? and Ka values?

Answer 1

The equation for pH is:

pH = -log [H+].

So, if you know the concentration of hydronium ions in the solution, you can calculate the pH.

For any weak acid, HA, the acid will dissociate in solution by the following equation:

HA + H2O --> H3O+ + A-

For a variety of reasons, the H2O is left out of the equation, and the reaction becomes:

HA --> H+ + A-

The equilibrium constant expression for this reaction is:

Ka = [H+][A-]/[HA]

In some problems, you will be given the concentration of the weak acid (and its Ka) and asked to calculate the pH of the solution. In that case, [H+] = [A-], so the equation becomes:

Ka = [H+]^2/[HA]

You can solve this equation for [H+] and calculate the pH from that.

There are many kind of problems that can be asked involving these equations.

Answer 2

first a note for the one who posted above me , writing H+ is wrong even if u take it that way at school , the right thing to write is H3O+ , cuz water dont give :
H2O ----------> H+ + HO- thats so wrong , its this way : 2H2O ----------> H3O+ + HO-



there is several cases :

first , if its the case of a strong acid u can calculate it according to : pH = -log Ca ( Ca is the concentration of the strong acid ) .

second , if its the case of strong base , u use this : pH = pKw + log Cb ( pKw = 14 , Cb is the concentration of the strong base )

third : if its the case of a week acid , u do the reaction of this acid with water : HA + H2O -------> A- + H3O+

u will be able to get the concentration of H3O+ , and then u calculate the pH as if H3O+ is a strong acid , as in the first case .

4th : if its a week base , u also use its reaction with water :
A- + H2O -------------> HA + HO-
then u can get the concentration of HO- and calculate the pH as if HO- is a strong base as in the second case .

5th : in the 4th and third cases u can calculate the pH if u have the concentrations of the couple of the week acid and its conjugate base or the week base and its conjugate acid , HA/A- , then u use , pH = pKa + log( [A-]/[HA] ) pKa is for the couple HA/A- ,.

now u can get the Ka if u have the pKa , then u say Ka = 10^(-pka) or if u dont have the pKa , and if u have the concentrations of the week acid and week base in the sloution u can calculate the Kr ( reaction constant ) and from it u can deduce the Ka , u get Kr by :
HA + H2O ----------------> A- + H3O+
Kr = [H3O+][A-]/[HA] = Ka , just in this case

A- + H2O -----------------> HA + HO-
Kr = [HA][HO-]/[A-] and now we multiply above and under by [H3O+] , so u get Kr = [HA][HO-][H3O+]/[A-][H3O+]
now [H3O+][HO-] = Kw = 10^(-pKw) = 10^(-14)
sp u get Kr=Kw/Ka since [A-]/[[HA][H3O+] = 1 / Ka
then u can easily get the Ka .

and in an easier way u can use that formula
" pH = pKa + log([A-]/[HA]) " only if u had the pH and if u have [A-] and [HA] so u substitue them in the equation and u get the pKa and then u get the Ka as i mentioned first .




hope i have answered ur question

Answer 3

the negative log of H conc is called pH.the dissociation constant of acidis called ka.