Question

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.

Answer 1

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.

Answer 2

The volume of 1.00 M sodium acetate solution needed to prepare an acetic/acetate buffer of pH 5.00 using 10.0 mL of 1.00M acetic acid solution is 17.8 mL.

We can find the volume of the acetate solution with the Henderson-Hasselbalch equation:

$pH = pka + log(\frac{[CH_{3}COONa]}{[CH_{3}COOH]})$   (1)  

Where:

[CH₃COOH] = 1.00 M                      

[CH₃COONa] =?

pH = 5.00

pKa = 4.75

From equation (1), we have:

$log(\frac{[CH_{3}COONa]}{[CH_{3}COOH]}) = pH - pKa$

$\frac{[CH_{3}COONa]}{[CH_{3}COOH]} = 10^{pH - pKa}$

$\frac{[CH_{3}COONa]}{[CH_{3}COOH]} = 10^{5.00 - 4.75} = 1.78$

Now, the volume of the acetate solution is:

$\frac{n_{CH_{3}COONa}/Vt}{n_{CH_{3}COOH}/Vt} = 1.78$  

Since the total volume is the same, we have:

$\frac{n_{CH_{3}COONa}}{n_{CH_{3}COOH}} = 1.78$  

$\frac{[CH_{3}COONa]_{i}*V_{b}}{[CH_{3}COOH]_{i}*Va} = 1.78$  

Solving for Vb

$Vb = \frac{1.78*[CH_{3}COOH]_{i}*Va}{[CH_{3}COONa]_{i}} = \frac{1.78*1.00M*10.0mL}{1.00 M} = 17.8 mL$

Therefore, we need to add 17.8 mL of sodium acetate solution.

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