Question

Consider the following data on some weak acids and weak bases:

acid Ka name formula nitrous acid HNO2 ×4.510−4 acetic acid HCH3CO2 ×1.810−5

base Kb name formula ammonia NH3 ×1.810−5 ethylamine C2H5NH2 ×6.410−4

Use this data to rank the following solutions in order of increasing pH. In other words, select a '

1

' next to the solution that will have the lowest pH, a '

2

' next to the solution that will have the next lowest pH, and so on.

solution pH 0.1 M NaNO2 choose one1 (lowest)234 (highest) 0.1 M NH4Cl choose one1 (lowest)234 (highest) 0.1 M NaI choose one1 (lowest)234 (highest) 0.1 M C2H5NH3Br

Conside - following data on some weak acids and weak bases: acid name formula nitrous acid HNO 4.5 x 10 base IK name formula ammonia NH; 1.810 ethylamine C,H, NH2 6.4 x 10-4 acetic acid HCH,CO2 1.8 x 10-5 Use this data to rank the following solutions in order of increasing pH. In other words, select a 'l' next to the solution that will have the lowest pH, a '2'nca to the solution that will have the next lowest pH, and so on. solution грн 0.1 M NaNO2 choose one 0,1 M NH4Cl choose one ! 0.1 M Nal choose one ! choose one : 0.1 MCHNH Br xs ?

Answer 1

solution pH rank 0.1 M NaNO2 0.1 M NH4Cl 1 0.1 M Nal 0.1 M C2H5NH3Br 2 NaNO2 is basic salt formed form strong base and weak acid HNO2 pka = -log[Ka] = -log[ 4.5x10-4 ] = 3.35 pH = [14 + pka + logc]/2 = [14+ 3.35 + log(0.1) 1/2 = 8.17 NH4Cl is acidic salt formed form strong acid and weak base NH3 pKb = -log[Kb] = -log[ 1.8x10-5 ] = 4.74 pH = [14-pKb-logC ]/2 = [ 14 - 4.74 - log(0.1) 1/2 = 5.13 Nal is neutral salt formed form strong acid and strong base. So pH = 7 C2H5NH3Br is acidic salt formed form strong acid and weak base C2H5NH2 pKb = -log[Kb] = -log[ 6.4x10-4 ] = 3.19 pH = 114-pKb -logC 1/2 = [ 14- 3.19 - log(0.1) 1/2 = 5.90