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What volume of 0.307 M NaOH must be added to produce a buffer of pH = 4.250?
What volume of 0.307 M NaOH must be added to 200.0 mL of 0.425 M Acetic Acid (Ka = 1.75 x 10^-5) to produce a buffer of pH = 4.250 ?
Thanks for your help!!
1 Answer
- deadfishfactoryLv 71 decade agoFavorite Answer
Ka = 1.75E-5, so the pKa = -log[ka] = 4.76
200 mL * 0.425 M = 85 mmoles acetic acid
Use the Henderson-Hasselbalch equation:
pH = pKa + log[A-/HA]
By adding NaOH, you could produce some conjugate base by using up conjugate acid, so:
HA = 85 mmoles - x
A- = x
4.250 = 4.76 + log[x/(85-x)]
x/85-x = 0.309
x = 26.27 - 0.309x
1.309x = 26.27
x = 20.07 mmoles CH3COONa
85 - 20.07 = 64.93 mmoles CH3COOH
So you need 64.93 mmoles CH3COOH and 20.07 mmoles CH3COONa. You can make this by adding the equivalent NaOH as you need CH3COONa.
So:
20.07 mmoles CH3COONa / 0.307 M NaOH = 65.37 mL NaOH