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

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

the correct answer is:

a. formation of precipitation