Use the Henderson Hasselbalch equation twice.
First set of conditions:
pH=9.0
pka=9.6
Solve for the ratio
pH=pka+log[A-/HA]
9.0=9.6+log[A-/HA]
10^(9.0-9.6)=[A-/HA]
0.25=[A-/HA]
Meaning 25% of the solution is protanated, or 0.025 moles is A- and 0.075 moles HA
Second set of conditions
pH=10.0
pka=9.6
Solve for the ratio
pH=pka+log[A-/HA]
10=9.6+log[A-/HA]
10^(10.0-9.6)=[A-/HA]
25.0=[A-/HA]
Meaning 25% of the solution is deprotanated, or 0.025 moles is HA and 0.075 moles A-
I need to go from 0.025 moles of A- to 0.075 moles of A-, and I need to go from 0.075 moles of HA to 0.025 moles of HA. This means that I need 0.05 moles of KOH.
5M KOH=0.05 moles of KOH/L
Solve for volume,
0.05 moles of KOH/5M=L of KOH
How much 5M KOH must be added to 1.0 L of 0.1M glycine (pKa 9.6) at pH 9.0 to bring its pH to exactly 10.0?
3 answers
Typos---
Change 0.025 to 0.25
change 0.075 to 0.75
change 0.05 to 0.5
Everything is correct.
Change 0.025 to 0.25
change 0.075 to 0.75
change 0.05 to 0.5
Everything is correct.
Never mind, I was correct the first time.