Suppose you want to make an acetic acid/acetate buffer to a pH of 5.00 using 10.0 mL of 1.00 M acetic acid solution. How many milliliters of 1.00 M sodium acetate solution would you need to add? The pKa for acetate buffer is 4.75.

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.

confused

confused             hewdejdnjnjsnd

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa