Answer:
Explanation:
Molarity of NaOAc needed
Using the Henderson-Hasselbalch Equation calculate base molarity needed given [HOAc] = 1.00M and pKa(NaOAc) = 4.75 and [HOAc] = 1.00m.
pH = pKa + log [NaOAc]/[HOAc]
5.00 = 4.75 + log[NaOAc]/[1.00M]
5.00 - 4.75 = log [NaOAc] - log[1.00M]
log [NaOAc] = 0.25 => [NaOAc] = 10⁰·²⁵ M = 1.78
Given 10ml of HOAc, how much (ml) 1.78M NaOAc to obtain a buffer pH of 5.00.
Determine Volume of Base Needed
(M·V)acid = (M·V)base => V(base) = (M·V)acid / (M)base
Vol (NaOAc) needed = (1.00M)(0.010L)/(1.78M) = 0.0056 liter = 5.6 ml.
Checking Results:
5.00 = 4.75 + log [1.78M]/[1.00M] = 4.75 + 0.25 = 5.00 QED.
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