Question

Given the following substances and their initial concentration:

a)	0.100 M HNO3	f)	0.100 M HF	k)	0.100 M HC6H5O
b)	55.5 M H2O	g)	0.100 M HNO2	l)	0.100 M Ba(OH)2
c)	0.100 M NaOH	h)	0.100 M CH3NH2	m)	0.00491 M HF
d)	0.100 M C2H5NH2	i)	0.100 M C5H5N	n)	0.100 M HOCl
e)	0.100 M KNO2	j)	0.100 M CH3NH3Br	o)	0.100 M CaCl2

Calculate the [H3O+] and the pH of each of the solutions.

Answer 1

a)

HNO3 is a strong acid, will form 100% disosciation

HNO3(aq) = H+(aq) + NO3-(aq)

then

[H+] = [HNO3] = 0.1 M due to stoichiometry

[H+] = [H3O+] = 0.1 M

then

pH = -log([H+])

pH = -log(0.1) = 1

this is very acidic, as expected since HNO3 is a very storng acid

b)

55.5 M of H2O

H2O is amphoteric, can act either as a base, accepting H+ to form H3O+ or cna act as an acid, donating H+ H2IO = H+ + OH-

in equilibrium, at T = 25 C, there is an equilbirium

Kw = [H3O+][OH-]

Kw = 10^-14 is constant

so

[H3O+] = sqrt(10^-14) = 10^-7 M

pH = -log(H3O+)

pH = -log(10^-7)

pH = 7

c)

C2H5NH2 is a weak base

will hydrolyse in water as follows

C2H5NH2 + H2O = C2H5NH3+ + OH-

the Kb expression is given by

Kb = [C2H5NH3+][OH-]/[C2H5NH2]

Kb = 4.3*10^-4 (constant value foundin books)

initially

[C2H5NH3+]= 0

[OH-] = 0

[C2H5NH2] = 0.1

in equilibrium

[C2H5NH3+]= 0 + x

[OH-] = 0 + x

[C2H5NH2] = 0.1 - x

then, substitute values:

Kb = [C2H5NH3+][OH-]/[C2H5NH2]

4.3*10^-4 = x*x/(0.1-x)

x^2 + (4.3*10^-4)x - (0.1)(4.3*10^-4) = 0

solve for x

x = 0.006345

then, [OH-] = x= 0.006345

pOH = -log(OH-) = -log(0.006345) = 2.19756

pH = !4-pOH = 14-2.19756

pH = 11.802

[H3O+] = 10^-pH = 10^11.802 = 1.577*10^-12 M

this is basic, as expected

d)

KNO2 is a salt, will ionize 100%

K+ + NO2- is present in water

NO2- is a conjugate base, so

NO2- + H2O = HNO2 + OH- is formed

this is basic

Kb = [HNO2][OH-]÷[NO2-]

Kb can be calcualted from Ka

Ka = 4*10^-4 for HNO2, so

Kb = (10^-14)/(4*10^-4) = 2.5E-11

initially

[HNO2] = 0

[OH-]= 0

[NO2-] = 0.1

in equilibrium

[HNO2] = 0 + x

[OH-]= 0 + x

[NO2-] = 0.1 - x

substitute in Kb

Kb = [HNO2][OH-]÷[NO2-]

2.5*10^-11 = x*x/(0.1-x)

x = 1.58*10^-6

x = Oh- =  1.58*10^-6

[H3O+] = (10^-14)/( 1.58*10^-6) = 6.33*10^-9 M