# Homework Help: PH Calculation

1. Dec 13, 2013

### Qube

1. The problem statement, all variables and given/known data

Consider mixing a 100 mL solution of 0.1 M sodium sulfide with a 200 mL solution of 0.05 M hydrochloric acid. What is the pH of the resulting solution?

2. Relevant equations

Moles of solute = molarity of solvent * volume of solvent

3. The attempt at a solution

Examining the first solution, we have 0.02 moles of sodium ion and 0.01 moles of sulfide ion. In solution, however, the sulfide ion is instantly protonated since it is a stronger base than hydroxide ion and according to the leveling effect any acid or base stronger than hydronium ion or hydroxide ion in a water based solution cannot exist in any large quantity.

Sulfide ion is therefore protonated to hydrogen sulfide ion, which is a much weaker base than hydroxide.

Examining the second solution, we have 0.01 moles of chloride ion, and 0.01 moles of hydrogen ion, which is instantly protonates water to form hydronium ion (HCl is a strong acid).

Neither the halogen nor the metal ion in this problem will affect pH to any significant extent, so we’ll ignore these. We’re only concerned about the acids and bases in the solutions of each of these substances. These are hydronium ion, hydroxide ion, hydrogen sulfide ion, and water.

The strongest acid and base – hydronium and hydroxide – will react to form water.
There are still two remaining species in solution – water and hydrogen sulfide ion. Water is amphiprotic – it can behave as either an acid or base. Hydrogen sulfide ion is similarly amphiprotic.

Consulting my acid-base table, I find that hydrogen sulfide ion is a weaker acid than water. So I’m not going to be worried about hydrogen sulfide ion behaving as an acid in water. Hydrogen sulfide, however, is a stronger base than water, so HS- will be the base, and H2O will be the acid.

Equation: HS- + H2O --> HO- + H2S (the conjugate base and acid, respectively, and no, I have not written the formula for hydroxide ion incorrectly – only contrary to tradition).

Since we are generating hydroxide ion in water and hydrogen sulfide in water, the solution will be basic, as hydroxide ion is a vastly stronger base than hydrogen sulfide is an acid.

Now we want to determine the pH of the reaction. We must first arrive at the base ionization constant of hydrogen sulfide ion, keeping in mind that:

Kw = Ka * Kb, where a and b are a conjugate acid-base pair. Ka refers to hydrogen sulfide; Kb would be referring to hydrogen sulfide ion. Kw = 1*10^-14. Ka = 1.1 * 10^-7

Kb = Kw / Ka = 9.09 * 10^-8.

We understand Kb to equal ([H2S][HO-])/[HS-] = x^2 / (mi – x), where x is negligible relative to mi, we can solve for x.

Kb (known) = x^2 / (mi – 0)

mi = 0.01 mole / 0.30 liters (combined solution volume)

x = 1.7 * 10^-5. This is the concentration of the conjugate acid, and the conjugate base, hydroxide ion.

-log[HO-] = 4.26 = pOH.

Since pH + pOH = 14, pH = 14 – 4.26 = 10 (to 2 significant figures).

This makes sense; we’re expecting the solution to be basic.

1) Still learning here. Does everything seem sound? No logical hole, flaws, errors? What about calculation errors? Any tips in general for pH calculations?

Last edited: Dec 13, 2013
2. Dec 13, 2013

### Staff: Mentor

I am just leaving for a funeral, so not much time. I just skimmed - nothing cries "I am wrong!!!"

But I would go much simpler route. There is a stoichiometric amount of H+ - just enough to protonate S2- to HS-. That means after mixing we have just a HS- solution (as if we were dissolving NaHS). And its pH can be calculated from a simplified formula

$$pH = \frac 1 2 (pK_{a1} + pK_{a2})$$

derived here.