salt hydrolysis and pH calculations

Question

I am a high school student and I am very confused in salt hydrolysis

My confusion is that: when we talk of hydrolysis of salt of weak base and weak acid, we internally assume the degree of hydrolysis of both cation and anion of the salt is assumed to be same but I think it is a very wrong assumption, why don't we simply do the pH calculation like I have done in this example? what is the problem with this?,if my approach is also correct then why doesn't it matches with the standard answer? what is the mistake in it? I have found some similar questions but no one has given the exact answer to those that's why I am asking it here please, explain it in a simplified way because I am only at high school level.

Answer 1

The general approach to solve acido-basic equilibrium problems is:

• respecting charge balance and charge conservation
• respecting mass conservation
• respecting dissociation equilibrium constants for acids, bases and water itself.

Generally, it leads to a set of nonlinear equations for a set of independent variables. The application of a substitution method usually results to a polynomial equation for $\ce{[H+]}$ of the second up to the forth order. The latter is hard to solve analytically.

Fortunately, in many scenarios, there can be made justified simplifications, like

• neglecting water autodissociation
• neglecting concentration of one substance form wrt the other one or their sum.

That largely simplifies an equation to be solved.

E.g. to calculate $\mathrm{pH}$ of solution of acetic acid, ammonia, or sodium carbonate, we can consider none of $\ce{H+}$, resp. $\ce{OH-}$ are created by the water autoionization and just negligible part of the compound underwent the acido-basic reaction.

E.g. for sodium acetate hydrolysis

$$K_\mathrm{b}=\frac{\ce{[HA][OH-]}}{\ce{[A-]}}\approx \frac{{\ce{[OH-]}}^2}{\ce{[A-]}}\approx \frac{{\ce{[OH-]}}^2}{c_0} $$

Answer 2

Regarding your degree of hydrolysis expression in the textbook, we have to keep charge balance in mind. It is not an assumption, it is a charge balance fundamental requirement because a solution can never have a uncharged specie without its counter ion. A solution is always electrically neutral. For example, $\ce{Na+}$ cannot exist alone in a solution, it also needs one singly charge anion like $\ce{Cl-}$ or $\ce{NO3-}$ to balance the charge. Same is the case of hydrolysis, if you lose one ammonium ion on the left hand side you must also lose a single formate ion.

Answer 3

LOL - I spent so much time trying to decipher your handwriting I missed what you question was.

You actually asked a very pertinent question. It is very very wise to wonder about the boundary conditions of some derived formula.

First let me point out that chemistry problems aren't like pure math problems. In math pi has been calculated to millions of decimal places. In chemistry 2-4 significant figures is typically the best that can be done.

You are absolutely right. For ammonium formate there are two hydrolysis reactions.

$$\ce{NH4+ + H2O <=> NH3 + H3O+}\tag{1}$$

$$\ce{HCOO- + H2O <=> HCOOH + OH-}\tag{2}$$

Now the solution must be electrically neutral so for the charge balance we have:

$$\ce{[NH4+] + [H+] = [HCOO-] + [OH-]}\tag{3}$$

However let's consider making a strong ammonium formate solution say 0.1 molar. The ammonium cation is a weak acid, and the formate anion is a weak base. Thus little of either will hydrolyze. So we can assume:

$$\ce{[NH4+] \gg [H+]}\tag{4}$$

and

$$\ce{[HCOO-] \gg [OH-]}\tag{5}$$

thus for charge balance we can assume:

$$\ce{[NH4+] = [HCOO-] }\tag{6}$$

This in turn implies that the majority of ammonia and formic acid in solution are not formed by the reactions (1) and (2) but rather by the reaction:

$$\ce{[NH4+] + [HCOO-] <=> [NH3] + [HCOOH]}\tag{7}$$

Thus the two ions from the salt are hydrolyzed by same amount.

Now if we have instead made a weak solution of ammonium formate, say $1.0\cdot 10^{-5}$ molar, then equations (4) and (5) wouldn't be valid and the individual hydrolysis reactions would have to be considered making the formula to calculate the pH much more complicated.