1a Using the log CpH diagram given find the pH of the follow

1a. Using the log C-pH diagram given, find the pH of the following solutions i, 10-1 M H3Cit water ii. 10-1 M Na2HCitwater ii. 0.05 M Na3Cit0.05 M NaH2Citwater You may find the equilibrium pH with either a charge balance or a proton condition Citric Acid logH3Cit logH2Cit- 1.logHCit2 -5 -6 -8 pH

Solution

i) log C=log (10^-1)=-1 ,from the given graph ,pH=1

for water,pH=7.0

pH=-log [H+],

so [H+](H3Cit)=10^-pH=10^-1

[H+]wter=10^-7

total [H+]=10^-7+10^-1=10^-1(approx)

pH=-log (10^-1)=1

ii)log C=log (10^-1)=-1 ,from the given graph ,pH=5.4

for water,pH=7.0

[H+](HCit2-)=10^-5.4=3.981*10^-6 M

pH=-log [H+],

[H+]wter=10^-7

total [H+]=10^-7+(3.981*10^-6 M)=10^-7+(39.81*10^-7 M)(approx)=40.81*10^-7M

pH=-log (40.81*10^-7M)=5.4

pH=5.4

iii) log C(0.05M cit3-)=-1.3,from the given graph ,pH=6.4

pH=-log [H+],

so [H+](cit3-)=10^-pH=10^-6.4=3.981*10^-7M

log C((0.05M H2cit-))=-1.3,from the given graph ,pH=3.2

[H+](H2Cit-)=10^-3.2=6.309*10^-4M

total [H+]=3.981*10^-7M+6.309*10^-4M=3.981*10^-7M+6309*10^-7M=6312.981*10^-7M

pH=-log (6312.981*10^-7)=3.2

pH=3.2