Question

I am doing a post lab for a acid and base buffers lab. I was asked to find the theoretical pH (what the pH should actually be) for each of my pH's in my data please help it is due tonight and that is the only question I cannot figure out. I just need one of the theoretical pHsd' one and I can do the rest.

Background:

30 mL of sodium dihydrogen phosphate and 20 mL k2HPO4 was added to buffer one

20 mL of sodium dihydrogen phosphate and 30 mL of k2HPO4 was added to buffer two

HCl .15 M

NaOH .15 M

.200 M K2HPO4

.200 M K2HPO4

Answer 1

Parts 1 and 2

We will need the pK_a value of sodium dihydrogen phosphate, NaH₂PO₄ which is 7.20.

Write down the ionization reaction.

H₂PO₄^- (aq) <======> H⁺ (aq) + HPO₄^- (aq); pK_a = 7.20

H₂PO₄^- is supplied by NaH₂PO₄ while K₂HPO₄ is the source of HPO₄^2-. You haven’t mentioned the concentration of H₂PO₄^- and HPO₄^2-, but I shall assume them to be 0.200 M each.

Prepare the following table.

Part 1	Millimoles of H2PO4- = (volume of H2PO4-)*(concentration of H2PO4-)	Millimoles of HPO42- = (volume of HPO42-)*(concentration of HPO42-)	Total volume of solution (mL)	Concentration of H2PO4- = (millimoles of H2PO4-)/(total volume of solution)	Concentration of HPO42- = (millimoles of HPO42-)/(total volume of solution)
	(30 mL)*(0.200 M)	(20 mL)*(0.200 M)	50	(30 mL)*(0.200 M)/(50 mL) = 0.12 M	(20 mL)*(0.200 M)/(50 mL) = 0.08 M
Part 2	(20 mL)*(0.200 M)	(30 mL)*(0.200 M)	50	(20 mL)*(0.200 M)/(50 mL) = 0.08 M	(30 mL)*(0.200 M)/(50 mL) = 0.12 M

Use the Henderson-Hasslebach equation:

pH = pK_a + log [HPO₄^2-]/[H₂PO₄^-]

Plug in values and get

Part 1: pH = 7.20 + log (0.08 M)/(0.12 M) = 7.20 + log (0.6667) = 7.20 + (-0.1761) = 7.0239 ≈ 7.02 (ans).

Part 2: pH = 7.20 + log (0.12 M)/(0.08 M) = 7.20 + log (1.50) = 7.20 + (0.1761) = 7.3761 ≈ 7.40 (ans).

Part 3:

HCl dissociates completely into protons; the dissociation equation is

HCl (aq) -------> H⁺ (aq) + Cl^- (aq)

As per the dissociation equation, [HCl] = [H⁺] = 0.15 M.

pH = -log [H⁺] = -log (0.15) = 0.8239 ≈ 0.82 (ans).

Part 4:

NaOH dissociates completely into OH^-; the dissociation equation is

NaOH (aq) --------> Na⁺ (aq) + OH^- (aq)

As per the dissociation equation, [NaOH] = [OH^-] = 0.15 M.

pOH = -log [OH^-] = -log (0.15) = 0.8239 and pH = 14 – pOH = 14 – 0.8239 = 13.1761 ≈ 13.18 (ans).