Question

What is the final pH of a solution obtained by mixing 300 ml of 0.4 M NH3 with 175 ml of 0.3 M HCl? (Kb = 1.8 x 10-5) Show all of your math steps. Do not leave us guessing as to how you got your final answer.

Answer 1

pH of the final solution = 9.15

Step-by-step explanation:

Equation of the reaction: HCl + NH₃ ----> NH₄Cl

Number of moles of NH₃ = molarity * volume (L)

= 0.4 M * (300/1000) * 1 L = 0.12 moles

Number of moles of HCl = molarity * volume (L)

= 0.3 M * (175/1000) * 1 L = 0.0525 moles

Since all he acid is used up in the reaction, number of moles of acid used up equals number of moles of NH₄Cl produced

Number moles of NH₄Cl produced = 0.0525 moles

Number of moles of base left unreacted = 0.12 - 0.0525 = 0.0675

pOH = pKb + log([salt]/[base])

pKb = -logKb

pOH = -log (1.8 * 10⁻⁵) + log (0.0525/0.06755)

pOh = 4.744 + 0.109

pOH = 4.853

pH = 14 - pOH

pH = 14 - 4.853

pH = 9.15

Therefore, pH of the final solution = 9.15