Question

To make a phosphate buffer at pH 6.82 starting with one liter of 20 mM phosphoric acid (H3PO4; pKs are of 2.15, 6.82, and 12.38), you could add _____ mL of ____ M KOH.

a) 500 mL of 10 M KOH
b) 1000 mL of 5 M KOH
c) 250 mL of 20 M KOH
d) 750 mL of 7.5 M KOH

Answer 1

To make a phosphate buffer at pH 6.82 starting with one liter of 20 mM phosphoric acid (H3PO4), you could add 750 mL of 7.5 M KOH. The option (D) is correct.

Step-by-step explanation:

To make a phosphate buffer at pH 6.82 starting with one liter of 20 mM phosphoric acid (H3PO4; pKs are 2.15, 6.82, and 12.38), you could add 750 mL of 7.5 M KOH.

To calculate the amount of KOH needed, we use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

For a pH of 6.82 and pKa of 6.82, [A-]/[HA] = 1, so we need an equal molar amount of base and acid.

The molarity (M) of phosphoric acid (H3PO4) is 20 mM, which is equal to 0.020 M. The molarity of KOH needed will also be 0.020 M, and since 7.5 M KOH is available, we can calculate the volume using the equation:

M1V1 = M2V2

(7.5 M)(V1) = (0.020 M)(1 L)

V1 = (0.020 M)(1 L) / (7.5 M) = 0.00267 L = 2.67 mL

you should add 750 mL of 7.5 M KOH to make the phosphate buffer at pH 6.82. Therefore, option (D) is correct.