Question

What is the pH of the solution that results from adding 27.7 mL of 0.12 M HCl to 25.9 mL of 0.41 M NH3 ? (K_a for ammonium ion is 5.6 x 10^-10 .)

pH =

Answer 1

We know that Molarity is the number of moles of solute present in one litre of solution and is denoted by M.

M = Number of moles of solute/Volume of solution in Litres

M = Number of moles of solute*1000/Volume of solution in mL

M = Weight of solute *1000/(Molar mass of solute * Volume of solution in mL)

Number of moles of solute = Molarity of solution* Volume of solution in mL/1000

Step-1: Determination of number of moles of NH3 and HCl

Number of moles of HCl in 27.7 mL of 0.12 M HCl is

= 27.7 mL * 0.12 M/1000

= 0.003324 moles

Number of moles of NH3 in 25.9 mL of0 41 M NH3 is

= 25.9 mL * 0.41 M/1000

= 0.01062 moles

Step-2: Determination of concentration in terms of Molarity of NH3 and NH4Cl after mixing HCl and Ammonia

The reaction between HCl and Ammonia (NH3) is

When 0.003324 moles of HCl is added to 0.01062 moles of NH3, 0.003324 moles of NH3 is neutralized and 0.003324 moles of NH4Cl is formed.

Number of moles of NH3 unreacted = 0.01062 moles - 0.003324 moles = 0.007295 moles

The resultant solution contains 0.003324 moles of NH4Cl and 0.007295 moles of NH3.

Since  27.7 mL of 0.12 M HCl is mixed with 25.9 mL 0.41 M NH3,

Total volume of the solution = 27.7 mL + 25.9 mL = 53.6 mL

According to the definition of molarity,

Concentration of NH3 and NH4Cl in the resultant solution is

[NH3] = 0.007295 moles * 1000/53.6 mL

= 0.136 M

Concentration of NH4Cl in the resultant solution is

[NH4Cl] = 0.003324 moles * 1000/53.6 mL

= 0.0620 M

Step-3: Determination of Kb and pKb

Given that Ka of NH3 is

K_a of Ammonium ion = 5.6 x 10^-10

We know that

We know that Kw = 1.0 * 10^-14

pKb = -log(Kb) = -log(1.78*10^-5)

pKb = 4.75

Step-4: Calculation of pOH & pH of buffer:

The mixture of Ammonia and Ammonium Chloride forms a basic buffer solution.

Buffer solution is a solution which resists the change in pH.

Buffer solutions are of two types.

1. Acidic buffer: It is a mixture of weak acid and salt formed by this weak acid with a strong base.

Ex: Acetic acid+ Sodium acetate

2. Basic buffer: It is a mixture of weak base and salt formed by this weak base with a strong acid.

Ex: Ammonium hydroxide (weak base) and Ammonium chloride (salt formed by Ammonium hydroxide with HCl)

pH of acidic buffer is given by

pH = pKa + log([Salt]/[Acid])

pOH of basic buffer is given by

pOH = pKb + log([Salt]/[Base])

For this resultant solution, pOH is given by

pOH = 4.4

We know that pH + pOH = 14

pH = 14 - pOH = 14 - 4.4 = 9.6