A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after 50.0 mL of the NaOH solution were added? The Kb of ammonia is 1.76×10^-5.

A. 9.25
B. 4.75
C. 11.22
D. 5.02
E. 8.98
**Edit: Answer is A, 9.25**

pH = 9.25

Explanation:

Step 1: Data given

Volume of a 0.40 M NH4Cl solution = 200.0 mL = 0.200 L

Molarity of NaOH = 0.80 M

Volume of NaOH = 50.0 mL

The Kb of ammonia is 1.76×10^-5

Step 2: The balanced equation

NH4Cl + NaOH → NH3 + NaCl + H2O

Step 3: Calculate moles

Moles NH4Cl  = molarity *¨volume

Moles NH4Cl = 0.40 M * 0.200 L = 0.08 moles

Moles NaOH = 0.80 M * 0.050 L = 0.04 moles

Step 4: Calculate limiting reactant

NaOH will completely be consumed. (0.04 moles). NH44Cl is in excess. There will react 0.04 moles .there will remain 0.04 moles.

Step 5: Calculate moles NH3

Moles NH3 = for 0.4 moles NH4Cl we'll have 0.4 moles NH3

Step 6: Calculate molarity

Molarity = moles / volume

Molarity = 0.4 moles = 0.250 L

Molarity = 1.6 M

Step 7: Calculate pOH

pOH = pKb + log ([B+]/[BOH])

pOH = 4.75 + log (1.6 / 1.6)

pOH = 4.75

Step 8: Calculate pH

pH = 14 - 4.75

pH = 9.25

i believe the answer is 5,000