Question

What is the pH of the buffer that results when 6.54 g of NH₃ and 5.35 g of NH₄Cl are diluted with water to a volume of 500 mL? (K_a for NH₄+ = 5.6 × 10-10.)

A.5.33

B.9.84

C.9.16

D.8.67

E.9.34

Answer 1

Answer

E) 9.34

Explanation

Given moles of NH3 = 6.54g /17.031g/mol = 0.3840

[NH3] = (0.3840mol/500ml)×1000ml = 0.7680M

given moles of NH4⁺ = 5.35g/18.039g/mol = 0.2966

[NH4⁺] = (0.2966mol/500ml)×1000ml = 0.5932M

Henderson-Hasselbalch equation is

pH = pKa + log([A^-] /[NH4⁺])

[A^-] = [NH3]

[HA] = [NH4⁺]

Ka of NH4⁺ = 5.6×10^-10

pKa = -logKa

pKa = - log(5.6×10^-10)

pKa = 9.25

substituting values in Henderson-Hasselbalch equation

pH = 9.25 + log(0.7680M/0.5932M)

pH = 9.25 + 0.11

pH = 9.36