Question

To make a phosphate buffer at pH 12.38 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) 10 mL of 0.1 M KOH
b) 100 mL of 1 M KOH
c) 1000 mL of 10 M KOH
d) 10000 mL of 100 M KOH

Answer 1

To create the desired phosphate buffer at pH 12.38, one must add stoichiometric amounts of KOH to phosphoric acid to reach a 1:1 ratio of conjugate base to acid. The provided options do not contain a feasible solution, as they suggest impractical or unavailable concentrations and volumes of KOH.

Step-by-step explanation:

To make a phosphate buffer at pH 12.38 starting with one liter of 20 mM phosphoric acid (H₃PO₄), with pKa values of 2.15, 6.82, and 12.38, you would want to convert the weak acid (H₃PO₄) to its conjugate base that corresponds to the third dissociation (since the third pKa is matching the desired pH of the buffer). This means you need to add enough base to deprotonate the acid to its third deprotonated form, which in this case is PO₄³-.

For a buffer at pH equal to the pKa, the ratio of the conjugate base to the acid is 1:1. Since we want the pH to equal the third pKa of phosphoric acid, we would need to add stoichiometric amounts of a strong base such as KOH to achieve the right amount of the conjugate base, creating a buffer. Considering the molarity and volume given, you would also have to account for dilution.

None of the provided options (a through d) are feasible because the highest concentration of KOH available is typically around 10 M, and even if higher concentrations were possible, the volumes are impractical. Therefore, the correct answer must be determined by calculating the actual amount of KOH required.