Question

What is the final pH of a solution with 50 mL of a stock solution of 2.0 M K2HPO4 and 25 mL of a stock solution of 2.0 M KH₂PO₄; diluted to a final volume of 200 mL. (pK=6.82; log(1) = 0; log(2)=0.3; log(3)=0.48; log(5)=0.7).

a) 7.0
b) 7.5
c) 8.0
d) 8.5

Answer 1

To calculate the final pH of the buffer solution composed of K2HPO4 and KH2PO4, the Henderson-Hasselbalch equation is used. The calculated pH is 7.12, which when rounded to the nearest 0.5 due to the level of precision of the given log values leads to a final pH of 7.0.

Step-by-step explanation:

We have a mixture of KH2PO4 and K2HPO4, which form a buffer system. To find the pH of a buffer, we use the Henderson-Hasselbalch equation: pH = pKa + log ([A-]/[HA]). We are given pKa (pK value) as 6.82. From the volumes and molarities given, we can calculate the moles of KH2PO4 (acid, HA) and K2HPO4 (base, A-).

Moles of K2HPO4 = 0.050 L * 2.0 M = 0.100 moles.
Moles of KH2PO4 = 0.025 L * 2.0 M = 0.050 moles.

After dilution, the total volume is 0.200 L, but the mole ratio doesn't change, so we can plug the values directly into the Henderson-Hasselbalch equation without adjusting for the final volume of the solution.

pH = 6.82 + log (0.100 moles/0.050 moles) = 6.82 + log(2) = 6.82 + 0.3 = 7.12

Because the options given don't include 7.12 and pH is typically reported to the nearest 0.5 when dealing with significant figures in the given log values, the closest option is 7.0.