A buffer solution contains 0.298 M ammonium chloride and 0.478 M ammonia. If 0.0560 moles of hydroiodic acid are added to 225 mL of this buffer, what is the pH of the resulting solution?

pH = 8.87

Explanation:

Hydroiodic acid, HI, is a strong acid that reacts with ammonia, NH3, to produce ammonium ion, NH⁴⁺. That means the moles of HI added = moles of NH3 consumed and moles of NH4⁺ produced.

Initial moles NH₄⁺:

0.225L * (0.298mol/L) = 0.06705 moles

Initial moles NH3:

0.225L * (0.478mol/L) = 0.10755 moles

After the reaction the moles are:

0.10755moles NH3 - 0.0560moles = 0.05155 moles NH3

0.06705moles NH4+ + 0.0560moles = 0.12305 moles NH4+

Using H-H equation for weak bases:

pOH = pKb + log ([NH4+] / [NH3])

pKb for ammonia is 4.75, [NH4+] could be the moles of NH4+ = 0.12305mol,

[NH3] = 0.05155moles

Replacing:

pOH = 4.75 + log (0.12305mol / 0.05155moles)

pOH = 5.13

pH = 14-pOH

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

the answer is d. 74.02