Calculate the logarithm of the ratio of the concentration of ammonia to ammonium

Question

Calculate the logarithm of the ratio of the concentration of ammonia to ammonium chloride, i.e., find log([base]/[acid]), for each of the solutions listed in Table 2. Be sure to use the actual concentrations of the solutions given to you in the laboratory. The concentration of Ammonia and Ammonium Chloride is 50.0 M/with the Kb of Ammonia being 5.6*10^-10. Please show work I have no idea how to do this calculation.

Solution Volume of ammonium chloride(mL) Volume of ammonia (mL) pH A1 15 15 9.69 B1 20 10 9.39 C1 25 5 9.04 D1 5 25 10.31 E1 10 20 9.72

Explanation / Answer

Using H-H equation,

pOH = pKb + log [conjugate acid]/[base]

NH3 + HCl-----> NH4Cl + H2O

so, given Kb = 5.6*10^-10.

pKb = -log Kb = 9.25

so pOH = pKb + log [NH4Cl]/[NH3]

Case A1

given pH = 9.69

so pOH = 14 - pH = 4.31

so, 4.31 = 9.25 + log [NH4Cl]/[NH3]

log [NH4Cl]/[NH3] = -4.94

log[NH3]/[NH4Cl] = 4.94

Case B1

given pH = 9.39

so pOH = 14 - pH = 4.61

so, 4.61 = 9.25 + log [NH4Cl]/[NH3]

log [NH4Cl]/[NH3] = -4.64

log[NH3]/[NH4Cl] = 4.64

Case C1

given pH = 9.04

so pOH = 14 - pH = 4.96

so, 4.96 = 9.25 + log [NH4Cl]/[NH3]

log [NH4Cl]/[NH3] = -4.29

log[NH3]/[NH4Cl] = 4.29

Case D1

given pH =10.31

so pOH = 14 - pH = 3.69

so, 3.69 = 9.25 + log [NH4Cl]/[NH3]

log [NH4Cl]/[NH3] = -5.56

log[NH3]/[NH4Cl] = 5.56

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