A chemist titrates 110.0 mL of a 0.5568 M ammonia (NH_3) solution with 0.4412 M HCl solution at 25 degree C. Calculate the pH at equivalence. The pK_b of ammonia is 4.75. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added. pH =

pH = 4.93

Explanation:

Step 1: Data given

Volume of a 0.5568 M ammonia NH3 solution = 110.0 mL = 0.110 L

Molarity of HCl = 0.4412 M

pKb ammonia = 4.75

Step 2: The balanced equation

NH3 + HCl ⇒ NH4+ + Cl-

Step 3: Calculate moles NH3

Moles NH3 =  molarity NH3 * volume

Moles NH3 = 0.5568 M * 0.110 L

Moles NH3 = 0.06125 moles

Step 3: Calculate moles HCl needed

For 0.06125 moles NH3 we need 0.06125 moles HCl

Step 4: Calculate volume HClFor 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4+

Volume HCl = moles HCl / molarity

Volume HCl = 0.06125 moles / 0.4412 M

Volume HCl = 0.139 L = 139 mL

Step 5: Calculate total volume

Total volume = 110.0 mL + 139 mL

Total volume = 249 mL

Step 6: Calculate moles NH4+

For 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4+

For 0.06125 moles NH3 we'll have 0.06125 moles NH4+

Step 7: Calculate [NH4+]

[NH4+] = moles NH4+ / volume

[NH4+] = 0.06125 moles / 0.249 L

[NH4+]= 0.246 M

Step 8: The equation

NH4+ + H2O → NH3 + H3O+

Step 9: The initial concentration

[NH4+] = 0.246 M

[NH3]= 0M

[H30+] = 0M

Step 10: Calculate the concentration at the equilibrium

[NH4+] = 0.246 - X M

[NH3]= X M

[H30+] = X M

Step 11: Calculate Ka

pKa = 14 - pKb

pKa = 14 - 4.75 = 9.25

Ka = 10^-9.25 = 5.6 *10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka = X² / (0.246 - X)

5.6 *10^-10 = X² / (0.246 - X)

5.6 *10^-10 = X² / (0.246)

X = 1.17*10^-5 M = [ H3O+]

Step 12: Calculate pH

pH = - log [1.17*10^-5]

pH = 4.93