Question

HC2HO3(aq)+H2O(l)⇄H3O+(aq)+C2HO3−(aq)
pKa=3.18
at 25°C

The equilibrium for the acid ionization of HC2HO3 is represented by the equation above. If 10.0mL of 0.20MHC2HO3 reacts with 5.0mL of 0.10MNaOH, which of the following could be used to calculate the correct pH of the resulting solution?

A) pH=pKa+log(0.100.20)

B) pH=pKa+log(0.0050×0.100.0100×0.20)

C) pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

D) pH=pKa+log[((0.0050×0.10)+(0.0100×0.20)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

Answer 1

Answer:

C) pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

Explanation:

To find the pH of a buffer (The mixture of the weak acid and its conjugate base) we have to use Henderson-Hasselbalch equation:

pH = pKa + log [Base] / [Acid]

Where pKa is the pka of the buffer and [] is molar concentration of the species of the buffer

The [Base] is equal to the concentration of NaOH added:

0.10M * (0.005L / 0.015L)

And the concentration of the acid [Acid] is the initial concentration of the acid - the concentration of the NaOH added:

0.0100L * (0.20M)/0.0150L - 0.10M * (0.005L / 0.015L)

That means the pH of the buffer is:

C) pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

Answer 2

Answer:

C) pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]

Explanation:

The equilibrium for the acid ionization of HC2HO3 is represented by the equation above. If 10.0mL of 0.20MHC2HO3 reacts with 5.0mL of 0.10MNaOH, which of the following could be used to calculate the correct pH of the resulting solution. pH=pKa+log[((0.0050×0.10)0.0150)/((0.0100×0.20)−(0.0050×0.10)0.0150)]