Water Chemistry 2nd ed. Chapter 8 problem 5 A solution contains only one acid/ba
ID: 1051359 • Letter: W
Question
Water Chemistry 2nd ed. Chapter 8 problem 5 A solution contains only one acid/base group, HA/A-, at a total concentration of 2×10-4 M. pKa for HA is 9.0. If the solution is titrated from pH 9.0 to 10.0, what fraction of the NaOH added is attributable to reactions of the TOTA? Put another way, what is the ratio of the amount of base is needed to carry out the titration with the HA/A- species present to the amount that would be required if they were absent? Water Chemistry 2nd ed. Chapter 8 problem 5 A solution contains only one acid/base group, HA/A-, at a total concentration of 2×10-4 M. pKa for HA is 9.0. If the solution is titrated from pH 9.0 to 10.0, what fraction of the NaOH added is attributable to reactions of the TOTA? Put another way, what is the ratio of the amount of base is needed to carry out the titration with the HA/A- species present to the amount that would be required if they were absent? Water Chemistry 2nd ed. Chapter 8 problem 5 A solution contains only one acid/base group, HA/A-, at a total concentration of 2×10-4 M. pKa for HA is 9.0. If the solution is titrated from pH 9.0 to 10.0, what fraction of the NaOH added is attributable to reactions of the TOTA? Put another way, what is the ratio of the amount of base is needed to carry out the titration with the HA/A- species present to the amount that would be required if they were absent?Explanation / Answer
You have to find the ratio between A¯ and HA so the concentrations are not needed
pH = pKa + log [base] / [acid] or pH = pKa + log [A¯] / [HA]
pH =9-10 , pKa =9 , log [base] / [acid] = ?
log [base] / [acid] = pH - pKa
= 10 - 9
log [A¯] / [HA] = 1
[A¯] / [HA] = 101
ratio of base form to acid form =10
= 10 to 1 (call it 1000 to 100)
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