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 issmall? (Ka for HF is 6.8×10−4.)

Part A

0.260 M

Part B

5.30×10⁻² M

Part C

2.60×10⁻² M

Express your answers to two decimal places.

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-
C 0 0 (initial)
C-x x x (at equilibrium)

Ka = x*x/(C-x)

A)
Ka = x*x/(C-x)
6.8*10^-4 = x^2 / (0.260-x)
1.768*10^-4 - 6.8*10^-4 * x =x^2
x^2 + 6.8*10^-4*x - 1.768*10^-4 = 0
solving above quadratic equation for positive value of x,
x = 0.013M

[H+] = 0.013 M

pH = -log [H+]
= -log (0.013)
= 1.89

B)
Ka = x*x/(C-x)
6.8*10^-4 = x^2 / (5.30*10^-2-x)
3.604*10^-5 - 6.8*10^-4 * x =x^2
x^2 + 6.8*10^-4*x - 3.604*10^-5 = 0
solving above quadratic equation for positive value of x,
x = 0.00567 M

[H+] = 0.00567 M

pH = -log [H+]
= -log (0.00567 )
= 2.25

C)
Ka = x*x/(C-x)
6.8*10^-4 = x^2 / (0.0260-x)
1.768*10^-5 - 6.8*10^-4 * x =x^2
x^2 + 6.8*10^-4*x - 1.768*10^-5 = 0
solving above quadratic equation for positive value of x,
x = 0.00388 M

[H+] = 0.00388 M

pH = -log [H+]
= -log (0.00388 )
= 2.41

D)
in (b) and (c) cocnentration is low and hence x can't be ignored