Question

Q(38) What is the pH of a 0.0200 M aqueous solution of HF (K. of HF is 6.8 A) pH=14 B) pH=2.4 C) pH 47 D) pH=9,3 10 ) E) pH=12,6 Q(39) What is the pH of a 0.1 M aqueous solution of NH3 (Ko of HF is 1.8 * 10-5)? A) pH=14 B) pH=2.4 C) pH 6.7 D) pH=11.3 6) pH=12.6 Q(40) Aqueous metal ions behave as Lewis acids because they A) can readily accept electrons in their vacant d orbitals. B) can readily donate electron pairs from their outermost p orbitals. C) attract electrons from the water molecules they are dissolved in. D)are attracted to the electronegative oxygen atoms in the solvent. E) they are also a little amphoter

Answer 1

38)

HF dissociates as:

HF -----> H+ + F-

2*10^-2 0 0

2*10^-2-x x x

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

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.8*10^-4)*2*10^-2) = 3.688*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.8*10^-4 = x^2/(2*10^-2-x)

1.36*10^-5 - 6.8*10^-4 *x = x^2

x^2 + 6.8*10^-4 *x-1.36*10^-5 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 6.8*10^-4

c = -1.36*10^-5

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 5.486*10^-5

roots are :

x = 3.363*10^-3 and x = -4.043*10^-3

since x can't be negative, the possible value of x is

x = 3.363*10^-3

So, [H+] = x = 3.363*10^-3 M

use:

pH = -log [H+]

= -log (3.363*10^-3)

= 2.4

Answer: B

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