Amino sulfonic acids is a strong acid with pKa=1.0. The pH=-log[H+]. Remember, [H+] is in mol/L.

Dear Andrei Blasko

I appreciate your reply

Actually I mean several compounds with a general name of " amino sulfonic acid", such as sulfamic acid and aminomethane sulfonic acid. You are right, in the case of sulfamic acid, known as amidosulfonic acid, the pKa is 1.01 and its pH is about 2.2, but I do not know how to find pH of some solutions of these compounds in which have pKa ranging from 5 to 11.

Do you have any suggestion?

best,

file:///C:/Users/arecktenwa/Desktop/biochemical%20calculation.pdf

Has a big section about pH.

Sorry, with this link you cannot do anything :)

http://site.iugaza.edu.ps/bzabut/wp-content/uploads/biochemical%20calculation.pdf

According to above post, you are interested in predicting the pH of sulfamic acid aq. sol., what can be achieved as follows:

Sulfamic acid is moderately strong; Ka = [(H2NSO3)-]·[H3O+]/[H3NSO3] = 0.101 M (pKa = 0.995) (*). Nonetheless, it is not advisable to be considered entirely dissociated.

Sulfamate molar balance writes: CH3NSO3 = [H3NSO3] + [(H2NSO3)-] = [(H2NSO3)-]·[H3O+]/Ka + [(H2NSO3)-] = [(H2NSO3)-]·{[H3O+] + Ka}/Ka, where CH3NSO3 is the nominal molar concentration of sulfamic acid (formality, to be precise).

Charge balance writes: [H3O+] = [OH-] + [(H2NSO3)-]. For sufficiently acid pH, [H3O+] > 100·[OH-] (i.e. pH < 6), [H3O+] ≈ [(H2NSO3)-].

Therefore: CH3NSO3 = [H3O+]·{[H3O+] + Ka}/Ka; or [H3O+] = - (½)(Ka) + {√{{CH3NSO3 + (¼)(Ka)}/M}·√(Ka/M)}M. Consequently:

[H3O+] = (½){ - 1 + √(1 + 4·CH3NSO3/Ka)}·Ka

pH = - log10{[H3O+]/M} = pKa - log10{(½){(√(1 + 4·CH3NSO3/Ka)) - 1}

Last expression can be generalized for aq. sol. of monoprotic acids (pH < 6) where the acid strength is of comparable order of magnitude to the acid nominal molar concentration (Cacid).

pH = pKa - log10{(½){(√(1 + 4·Cacid/Ka)) - 1}}

This expression is not valid for truly strong acids, which, by definition, are supposed to be entirely dissociated. Yet, also holds for weak monoprotic acids (pH < 6). It relates with the golden ratio for Cacid = Ka, provided that the acid solubility is not exceeded.

The following simplified equation can be derived from the last, for 4·Cacid/Ka > 100:

pH = pKa - log10{(√(Cacid/Ka)) - ½}

For Cacid/Ka > 2.5·103, last equation can be further simplified to:

pH = (½){pKa - log10(Cacid/M)}

Last two simplified equations may possibly apply to either sufficiently conc. aq. sol. of moderately strong acids, or to enough dilute weak acids, but not to really strong acids.

(*) https://en.wikipedia.org/wiki/Sulfamic_acid

For your inquiry of acids with pKa ranging from 5 to 11, they are weak acids and obey the Henderson-Hasselbalch equation:

pH=pKa+log([base]/[acid]) in which [base] is the concentration of the deprotonated acid (A-) in mol/L and [acid] is the protonated one (AH).

Dear @Achim Recktenwald and @Andrei Blasko

thank you for your valuable answers

Dear Carlos Araújo Queiroz

I appreciate your comprehensive answer. Thanks a million