Please calculate the pH of an aqueous 1mM solution of:?

Question

1)Imidiazole ,a base with pKb=7.01
2)1-NAPHTOL with pKa=9.34
3)morpholine,a base with pKb=8.5
4)HClO2 with pKa=1.95
5)HSCN with pKa =.090

Answer 1

1) Imidazole pH= 8.99
2) 1-Naphthol pH=6.17 or 6.14 (see below)
3) morpholine pH=8.24 or 8.25 (see below)
4) HClO2 pH=3.03
5) HSCN pH=3.00

How to calculate:
All your acids are monoprotic. Assume for these monoprotic acids the general reaction scheme:

.. .. .. .. .. .. HA <=> H(+) + A(-)
Initial .. .. .. .C
Dissociate.. x
Produce .. .. .. .. .. .. x.. .. .. .. x
At Equil.. .. C-x .. .. ..x .. .. .. ..x

Ka=[H+][A-]/HA] = x^2/(C-x) =>
x^2 + Kax -KaC=0
The only acceptable solution is
x=0.5*(-Ka + SquareRoot( Ka^2 +4KaC))
and pH=-logx

In your case C=10^-3 and Ka=10^-pKa

Actually when x is in the order of 10^-7 it is close to the self dissociation of water and then you should actually take into account water as well. the difference in your equation will be that you will also have 10^-7 initial [H+] so at equilibrium [H+]= x+10^-7
your equation becomes

x^2 + (Ka+10^-7)x -KaC=0 =>

x=0.5*(-(Ka+10^-7) + SquareRoot( (Ka+10^-7)^2 +4KaC))

and of course then pH=-log (x+10^-7)

Applying this correction for 1-naphthol you get pH= 6.14. For the other acids, Ka and C are big enough and you can overlook the dissociation of water.

Your bases are monoprotic and nitrogenous So you can assume the reaction scheme

B + H2O <=> BH(+) + OH(-) and as above (since H2O is never written in Kb expressions) you have

Kb= x^2/(C-x) =>
x=0.5*(-Kb + SquareRoot( Kb^2 +4KbC))
Then pOH=-logx and pH=14-pOH
You can do a similar correction for the water self-dissociation if you like for the case of morpholine and get pH=8.25 though the difference is neglegible.

I hope I haven't done any numerical errors...

Answer 2

Values of pH for weak and strong acids can be approximated using certain assumptions.

Under the Brønsted-Lowry theory, stronger or weaker acids are a relative concept. But here we define a strong acid as a species which is a much stronger acid than the hydronium (H3O+) ion. In that case the dissociation reaction (strictly HX+H2O↔H3O++X− but simplified as HX↔H++X−) goes to completion, i.e. no unreacted acid remains in solution. Dissolving the strong acid HCl in water can therefore be expressed:

HCl(aq) → H+ + Cl−
This means that in a 0.01 mol/L solution of HCl it is approximated that there is a concentration of 0.01 mol/L dissolved hydrogen ions. From above, the pH is: pH = −log10 [H+]:

pH = −log (0.01)
which equals 2.