Question

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

acid Ka name formula acetic acid HCH3CO2 ×1.810?5 hydrocyanic acid HCN ×4.910?10

base Kb name formula pyridine C5H5N ×1.710?9 aniline C6H5NH2 ×4.310?10

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 NaCl

0.1 M C6H5NH3Cl

0.1 M C5H5NHBr

0.1 M NaCN

Answer 1

1) 0.1 M NaCl

NaCl is a salt of strong acid HCl and strong base NaOH, so it forms neutral salt solution as the neutral salt NaCl is totally dissociated in solution.

pH(for neutral solution)7

2) 0.1 M C6H5NH3Cl

This is a salt of HCl(strong acid) and aniline(C6H5NH2) ,a weak base,so the solution is acidic

pH1/2(pKw-pKb-log C)

Kw1*10^-14,pKw-log Kw-log (1*10^-14)14

Kb4.310*10^-10,pKb-logKb-log (4.310*10^-10)9.365

pH1/2(14-9.365-log 0.1)2.817

pH2.82

3) 0.1M C5H5NHBr

This is a salt of HBr(strong acid) and pyridine (C5H5N) ,a weak base,so the solution is acidic

pH1/2(pKw-pKb-log C)

Kw1*10^-14,pKw-log Kw-log (1*10^-14)14

Kb1.710*10^-9,pKb-logKb-log (1.710*10^-9)8.77

pH1/2(14-8.77-log 0.1)3.11

pH3.11

4) 0.1M NaCN

This is a salt of weak acid HCN (hydrocyanic acid) and strong base NaOH.The solution is basic in nature with its pH given by the equation:

pH1/2(pKw+pKa+log C)

Kw1*10^-14,pKw-log Kw-log (1*10^-14)14.0

Ka4.910*10^-10,pKb-logKb-log (4.910*10^-10)9.31

pH1/2(14+9.31+log 0.1)11.1

pH11.1

solutions arranged in increasing order of pH:

1) 0.1 M C6H5NH3Cl

2) 0.1M C5H5NHBr

3) 0.1 M NaCl

4)0.1M NaCN