Crunching through K_a problems...

4. What is the expected pH of a 2.49M solution of acetic acid {K_a = 1.8x10^-5}? What is the expected pH when 100.0mL of this solution is combined with 100.0mL of water? Assume volumes are additive.

This is a K_a-type equilibrium problem, organize the information using a table.

	HC2H3O2(aq) +	H2O(l) ⇄	H3O+(aq) +	C2H3O2-1(aq)
[ ]initial	2.49 M	XXXX	0 M	0 M
Δ[ ]	- x	XXXX	+ x	+ x
[ ]equilibrium	(2.49 – x) M	XXXX	x M	x M

Assume “x” is much smaller than 2.49, plug in to the equilibrium constant expression...

x = 6.69x10^-3 = [H₃O⁺¹]

Assumption is good.

pH = -log[H₃O⁺¹] = -log(6.69x10^-3) = 2.174

For the second part, the set-up is the same, the only difference is that the initial concentration of acetic acid has been diluted. Calculating the dilution...

C₁V₁ = C₂V₂

(2.49M)(100.0mL) = C₂(200.0mL)

C₂ = 1.245M

Plug in and solve the same way:

x = 4.73x10^-3 = [H₃O⁺¹]

Assumption is still good.

pH = -log[H₃O⁺¹] = -log(4.73x10^-3) = 2.325

5. A 1.83M solution of a weak, monoprotic acid {HA(aq)} has a pH of 3.48. What is the K_a of this acid?

We can approach this as a K_a-type equilibrium problem as well, organize the information using a table.

	HA(aq) +	H2O(l) ⇄	H3O+(aq) +	A-1(aq)
[ ]initial	1.83 M	XXXX	0 M	0 M
Δ[ ]	- x	XXXX	+ x	+ x
[ ]equilibrium	(1.83 – x) M	XXXX	x M	x M

In this case, we are given a pH, which gives us a way to calculate [H₃O⁺], which gives us “x”...

[H₃O⁺] = 10^-pH = 10^-3.48 = 3.3113x10^-4 (Assumption is good.)

K_a = (3.3113x10^-4)(3.3113x10^-4) / (1.83) = 5.99x10^-8

6. You have combined 100.0mL of 2.84M hydrofluoric acid {K_a = 6.8x10^-4} and 100.0mL of 2.19M fluoride ions. What is the expected pH of the resulting solution? Assume volumes are additive.

Again, we can approach this as a K_a-type equilibrium problem (Noticing a pattern here?), organize the information using a table.

	HF(aq) +	H2O(l) ⇄	H3O+(aq) +	F-1(aq)
[ ]initial	1.42 M	XXXX	0 M	1.095 M
Δ[ ]	- x	XXXX	+ x	+ x
[ ]equilibrium	(1.42 – x) M	XXXX	x M	(1.095 + x) M

A couple little adjustments in this case... the original concentrations given in the problem have to be diluted to get the “initial” concentrations in the table. The other key difference here is that we're starting out with a mixture of reactants and products. That might clutter up the math a little bit, but it doesn't really change the way we approach the problem. To simplify things, let's assume that “x” is small compared to both 1.42 and 1.095. Then the K_a expression is:

6.8x10^-4 = (x)(1.095) / (1.42)

x = 8.818x10^-4

Assumption is good.

pH = -log[H₃O⁺¹] = -log(8.818x10^-4) = 3.055