Assuming no volume change on mixing,what mass of ammonium chloride should be added to 250.0 mL of 0.25 M aqueous ammonia to produce a solution of pH 10.70? (K_b for ammonia is 1.8 × 10^-5) A)3.7 mg B)120 mg C)40 mg D)30 mg E)80 mg

Question

Quizplus · Accepted Answer

First, we need to use the pKa of ammonium ion to calculate the concentration of ammonium ion at pH 10.7. The pKa of ammonium ion is 9.24 (derived from the given pKb value of ammonia). Therefore, at pH 10.7, the concentration of ammonium ion ([NH4+]) is:
[NH4+] = [NH3]/(1 + 10^(pKa-pH)) = 0.25/(1 + 10^(9.24-10.7)) = 0.0020 M
Next, we can use the balanced equation of the reaction between ammonium chloride and ammonia:
NH4Cl + NH3 → NH4+ + Cl-
to calculate the amount of ammonium chloride needed to react with all of the ammonium ion in the solution:
1 mole of NH4Cl reacts with 1 mole of NH4+
Therefore, the amount of NH4Cl needed is:
0.0020 moles NH4+ × 53.49 g/mol NH4Cl = 0.107 g NH4Cl = 107 mg NH4Cl
Therefore, the mass of ammonium chloride needed to add to 250.0 mL of 0.25 M aqueous ammonia is 107 mg, which is closest to answer choice B (120 mg).

Assuming No Volume Change on Mixing,what Mass of Ammonium Chloride