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?

Answer 1

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

Answer 2

Typos---

Change 0.025 to 0.25
change 0.075 to 0.75
change 0.05 to 0.5

Everything is correct.

Answer 3

Never mind, I was correct the first time.