Question

Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5, and Titrate with 0.100 mol HCl, calculate PH at 5 mL HCl, 10 mL , 15 mL, 20 mL, 24.5mL, 25mL, 25.5mL, 30mL 40.0 mL, abd 50 mL HCl. Graph the solution with hoizontal axis being the PH up to 14 and The y-axis being mL HCl.

Answer 1

pH of 12.5 mL of 0.2 M aqueous ammonia:

NH₃(aq)       ↔ NH₄⁺(aq) + OH^-(aq)
0.2-x M              x M             x M

K_b = [NH₄⁺] [OH^-] / [NH₃]

1.8 x10^-5 = (x)(x) / (0.2 - x)
Since the amount of x is very small compared to 0.2 M, it is neglected.
1.8 x10^-5 = (x)(x) / (0.2)
x = [OH^-] = 1.89 x 10^-3M

pOH = -log (1.89 x 10^-3) = 2.72

pH = 14 – 2.72 = 11.28

pH during the titration of 12.5 mL of 0.2 M aqueous ammonia with 5mL of 0.1 M HCl:

moles of aqueous ammonia = 0.0125 L x (0.2 mol/L) = 0.0025 moles

moles of HCl = 0.005 L x (0.1 mol/L) = 0.0005 moles

therefore [OH^-] = (0.0025-0.0005) moles/0.0175L = 0.114 M

pOH = -log (0.114) = 0.943

pH = 14 – 0.943 = 13.05

pH during the titration of 12.5 mL of 0.2 M aqueous ammonia with 10.00 mL of 0.1 M HCl:

moles of aqueous ammonia = 0.0125 L x (0.2 mol/L) = 0.0025 moles

moles of HCl = 0.010 L x (0.1 mol/L) = 0.001 moles

therefore [OH^-] = (0.0025-0.001) moles/0.0225L = 0.066M

pOH = -log (0.066) = 1.18

pH = 14 – 1.18 = 12.82

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 15.00 mL of 0.1 M HCl:

moles of aqueous ammonia = 0.0125 L x (0.2 mol/L) = 0.0025 moles

moles of HCl = 0.015 L x (0.1 mol/L) = 0.0015 moles

therefore [OH^-] = (0.0025-0.0015) moles/0.0275L = 0.03636M

pOH = -log (0.03636) = 1.43

pH = 14 – 1.43= 12.57

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 20 mL of 0.1 M HCl:

moles of aqueous ammonia = 0.0125 L x (0.2 mol/L) = 0.0025 moles

moles of HCl = 0.020 L x (0.1 mol/L) = 0.002 moles

therefore [OH^-] = (0.0025-0.002) moles/0.0325L = 0.0153M

pOH = -log (0.0153) = 1.81

pH = 14 – 1.81= 12.19

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 24.5 mL of 0.1 M HCl:

moles of aqueous ammonia = 0.0125 L x (0.2 mol/L) = 0.0025 moles

moles of HCl = 0.0245 L x (0.1 mol/L) = 0.00245 moles

therefore [OH^-] = (0.0025-0.00245) moles/0.0370L = 0.00135M

pOH = -log (0.00135) = 2.86

pH = 14 – 2.86= 11.14

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 25.00 mL of 0.1 M HCl:

moles of aqueous ammonia = 0.0125 L x (0.2 mol/L) = 0.0025 moles

moles of HCl = 0.025 L x (0.1 mol/L) = 0.0025 moles

therefore [OH^-] = (0.0025-0.0025) = zero moles NH₃ remaining

[NH₄⁺] = 0.0025 moles produced

This solution is no longer a buffer. It is now the salt of a weak base and its solution will be acidic. The actual molarity is important and it is 0.0666 M (0.0025 mol / 0.0375 L ). The pH is 1.18.

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 25.5 mL of 0.1 M HCl:

This solution is now a solution of a strong acid (the presence of the weak acid NH₄⁺ may be ignored).

Let us determine the new molarity of the HCl:

M₁V₁ = M₂V₂

(0.1 mol/L) (0.0005 L) = (x) (0.0380 L)

x = [H⁺]= 0.00131 M

pH = -log (0.00131) = 2.88

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 30 mL of 0.1 M HCl:

M₁V₁ = M₂V₂

(0.1 mol/L) (0.005 L) = (x) (0.0425 L)

x = [H⁺]= 0.0117 M

pH = -log (0.0117) = 1.93

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 40 mL of 0.1 M HCl:

M₁V₁ = M₂V₂

(0.1 mol/L) (0.015 L) = (x) (0.0525 L)

x = [H⁺]= 0.0285 M

pH = -log (0.0285) = 1.54

pH during the titration of 12.5mL of 0.2 M aqueous ammonia with 50 mL of 0.1 M HCl:

M₁V₁ = M₂V₂

(0.1 mol/L) (0.025 L) = (x) (0.0625 L)

x = [H⁺]= 0.04 M

pH = -log (0.04) = 1.39