Question

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (Ka for HF is 6.8×10−4.) a) 0.300 M b) 4.70×10⁻² M c) 2.60×10⁻² M

d) In which cases can you not make the simplifying assumption that x is small?

only in (a)

only in (b)

in (a) and (b)

in (b) and (c)

Answer 1

HF <==> H+ + F-

Ka = [H+][F-]/[HF]

	HF	H+	F-
initial	0.3	0	0
change	-x	+x	+x
equilibrium	0.3-x	+x	+x

Ka = x^2/(0.3-x)

or, 6.8×10−4 = x^2/(0.3-x)

as x is samll, 0.3-x = 0.3

x^2 = 6.8×10−4 * 0.3 = 0.014 M

pH= -log[H+] = 1.85

-------------------------------------------

Ka = x^2/(4.70 *10^-2-x)

or, 6.8×10−4 = x^2/(4.70 *10^-2-x)

or, x = 0.0053

pH = -log[H+] = -log[0.0053] = 2.28

---------------------------------------------------------

Ka = x^2/(4.70 *10^-2-x)

or, 6.8×10−4 = x^2/(2.6 *10^-2-x)

or, x = 0.0038

pH = -log[H+] = -log[0.0038] = 2.42

x can be neglected only when concentration is very high.