Calculate the pH of solution made by combining 75.0 mL of 0.10 M formic acid with 30.0 mL of 0.25 M KOH. (Ka for formic acid, HCOOH, is 1.8 X 10^-4)
answered above.
To calculate the pH of the solution, we need to determine the concentration of the resulting solution after the formic acid (HCOOH) and KOH reaction.
First, let's find the moles of each substance. We will use the formula:
moles = concentration (M) x volume (L)
For formic acid (HCOOH):
moles of HCOOH = 0.10 M x 0.075 L = 0.0075 mol
For potassium hydroxide (KOH):
moles of KOH = 0.25 M x 0.030 L = 0.0075 mol
Note that the coefficients in the balanced equation between HCOOH and KOH are both 1, so the number of moles of HCOOH and KOH is equal.
Since KOH is a strong base and HCOOH is a weak acid, KOH will react completely with HCOOH, resulting in the formation of water (H2O) and potassium formate (HCOOK). The equation would be:
HCOOH + KOH β H2O + HCOOK
Since the moles of KOH and HCOOH are equal, they will react completely. Therefore, there will be no excess KOH or HCOOH present.
The moles of HCOOH consumed and the moles of HCOOK formed are both 0.0075 mol.
Now, we will use the Henderson-Hasselbalch equation to calculate the pH. The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
In this case, [A-] refers to the formate ion (HCOO-) concentration, and [HA] refers to the formic acid (HCOOH) concentration.
Given that the Ka for formic acid (HCOOH) is 1.8 x 10^-4, the pKa is calculated as:
pKa = -log10(Ka) = -log10(1.8 x 10^-4) = 3.74
Now we need to calculate the concentration of formate ion (HCOO-) and formic acid (HCOOH).
The total volume of the resulting solution is the sum of the initial volumes of both solutions: 75.0 mL + 30.0 mL = 105.0 mL = 0.105 L.
The concentration of formate ion (HCOO-) is determined by dividing the moles of HCOOK by the total volume:
[HCOO-] = moles of HCOOK / total volume (L) = 0.0075 mol / 0.105 L β 0.0714 M
Similarly, the concentration of formic acid (HCOOH) is determined by dividing the moles of HCOOH by the total volume:
[HCOOH] = moles of HCOOH / total volume (L) = 0.0075 mol / 0.105 L β 0.0714 M
Now we can substitute the values into the Henderson-Hasselbalch equation:
pH = 3.74 + log ([0.0714] / [0.0714])
Simplifying,
pH = 3.74 + log(1)
Since the log of 1 is 0, the pH will be:
pH = 3.74 + 0 = 3.74
Therefore, the pH of the resulting solution is approximately 3.74.