Question

1. Calculate the equilibrium concentrations of all species present and the pH of a solution obtained by adding 0.100 moles of solid NaOH to 1.00 L of 15.0 M NH3. Kb = 1.8 × 10–5

2. One mole of a weak acid HA was dissolved in 2.0 L of water. After the system had come to equilibrium, the concentration of HA was found to be 0.45 M. Calculate the Ka for this weak acid.

3. Calculate the pH of a solution formed by mixing 50.0 mL of 0.40 M NH3 solution with 50.0 mL of a 0.40 M HCl solution. Is the resulting solution acidic, basic, or neutral? Ka = 5.6 × 10^–10

4. Which of the following mixtures would result in buffered solutions when 1.0 L of each of the two solutions is combined? Circle the letter of the appropriate response(s).

a. 1.0 M HBr and 1.0 M NaBr

b. 2.0 M HF and 1.0 M NaOH

c. 1.0 M HF and 1.0 M NaOH

d. 1.0 M NH3 and 1.0 M NH4Cl

e. 1.0 M NH3 and 0.50 M HCl

f. 1.0 M HNO3 and 0.50 M NaOH

5. How many grams of solid NaOH must be added to 1.00 L of 2.0 M CH3COOH (acetic acid) to produce a solution that is buffered at each pH? Ka = 1.76 × 10–5 a. pH = pKa b. pH = 4.00

Answer 1

1.

Molar concentration of NaOH = (0.1/1) = 0.1 M

Hence, concentration of added OH^- = 0.1 M

Concentration of NH₃ = 15.0 M

Ammonia solution forms the following equilibrium.

NH₃ + H₂O NH₄⁺ + OH^-

Using ICE table

	NH3	NH4+	OH-
Initial	15.0	0	0.1
Change	-x	+x	+x
Equilibrium	(15.0-x)	x	(0.1+x)

Kb = [NH₄⁺][OH^-] /[NH₃]

Or, Kb = x * (0.1+x)/(15 - x)

Or, 1.8*10^-5 = x(0.1+x)/(15-x)

Ammonia is weak base hence, degree of dissociation (x) is very low. So, 15 - x = 15.

Or, x(0.1+x)/15 = 1.8*10^-5

Or, 0.1x + x² = 15*1.8*10^-5

Or, x² + 0.1 x - 2.7*10^-4 = 0 (1)

Comparing with

ax² + bx + c = 0

a = 1, b= 0.1 , c = -2.7*10^-4

Then, using , x =

Then .Solve for equation 1

X =

= (0.105 - 0.1)/2 ( only positive value is taken , as x is alwasys positive)

Or, x = 0.00263

Then, equilibrium concentration of hydroxide ion ;

[OH^-] = 0.1+x = 0.1+0.00263 = 0.10263 M

Then, pOH = - log[0.100263) = 0.988

Then, pH = 14 - 0.988= 13.01

2.

[HA] = 1/2= 0.5 M

Final concentration = 0.45 M

Using ice table

	HA	H+	A-
Initial	0.5	0	0
Change	0.5-0.45	+0.05	+0.05
Equilibrium	0.45	0.05	0.05

Ka = [H⁺][A^-]/[HA] = (0.05*0.05)/0.45

Or, Ka = (0.0025/0.45)

Or, Ka = 0.00556

Hence , Ka of weak acid = 5.56*10^-3

3.

50 ml 0.40 M NH₃ completely reacts with 50 ml 0.40 M HCl.

Milimoles of salt = 50*0.4= 20

Total volume= (50+50)= 100 mL

Hence, concentration of salt ( NH₄Cl) = (20/200)= 0.2

Ka = 5.6*10^-10 , then Kb = 10^-14/5.6*10^-10 = 1.78*10^-5

Then pKb = -log(1.78*10^-5) = 4.75

Salt of weak base strong acid is

pH = 7 - ( pKb + logc)

pH = 7 - [4.75 + log(0.2)]

pH = (7 - 2.025)

pH = 4.974

4.

Buffer solution is solution of weak acid and its salt or weak acid and its salt.

a. HBr is strong acid, hence this solution is nit a buffer.

b. HF is weak acid , incomplete reaction with NaOH to form NaF .hence the solution contain both HF and its salt hence it is buffer.

C. HF and NaOH in complete reaction forms only salt.hence it is not buffer.

d. Equal molar ammonia ( weak base )and its salt NH₄Cl forms a buffer solution.

e) incomplete reaction of NH₃ with HCl forms salt . Solution contains excess ammonia and its salt. So the solution is buffer.

f)

HNO₃and NaOH both are strong.So the solution is not a buffer.