Respuesta :
Answer:
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
The pH of the solution after 50.0 mL of the NaOH solution were added should be considered as the 9.25.
Calculation of pH:
Since
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
Now moles is
Moles NH4Cl = molarity *¨volume
= 0.40 M * 0.200 L
= 0.08 moles
Moles NaOH = 0.80 M * 0.050 L = 0.04 moles
Now molarity is
Molarity = moles / volume
= 0.4 moles = 0.250 L
= 1.6 M
Now pOH
pOH = pKb + log ([B+]/[BOH])
= 4.75 + log (1.6 / 1.6)
= 4.75
And, finally pH
pH = 14 - 4.75
pH = 9.25
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