A solution is prepared by adding 100 mL of 0.2 M hydrochloric acid to 100 mL of 0.4 M sodium formate. Is this a buffer solution, and if so, what is its pH? A)It is a buffer, pH pK _a of formic acid. B)It is a buffer, pH

Question

A solution is prepared by adding 100 mL of 0.2 M hydrochloric acid to 100 mL of 0.4 M sodium formate. Is this a buffer solution, and if so, what is its pH? A)It is a buffer, pH > pK _a of formic acid. B)It is a buffer, pH < pK _a of formic acid. C)It is a buffer, pH = pK _a of formic acid. D)It is a buffer, pH = pK _b of sodium formate. E)Since hydrochloric acid is a strong acid, this is not a buffer.

Quizplus · Accepted Answer

The solution is a buffer because it contains a weak acid (formic acid) and its conjugate base (sodium formate). The pH of the buffer can be calculated using the Henderson-Hasselbalch equation:pH = pK _a + log([A-]/[HA])where pK _a is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.In this case, the pK _a of formic acid is 3.75. The concentration of the conjugate base is 0.1 M (since 100 mL of 0.4 M sodium formate was added to 100 mL of water), and the concentration of the weak acid is 0.1 M (since 100 mL of 0.2 M hydrochloric acid was added to 100 mL of water).Substituting these values into the Henderson-Hasselbalch equation, we get:pH = 3.75 + log(0.1/0.1) = 3.75Therefore, the pH of the buffer is 3.75, which is equal to the pK _a of formic acid.

A Solution Is Prepared by Adding 100 ML of 0