Calculate the pH of a solution that results from the mixture of 50.0 mL of 0.15 M HCOOH with 75.0 mL of 0.13 M HCOONa. Now, don't forget these solutions are being mixed and you haven't accounted for that yet in the concentrations! Ka for HCOOH is 1.8 x 10-4.
moles HCOOH = M x L = ?
moles HCOONa = M x L = ?
(HCOOH) = moles/total volume.
(HCOONa) = moles/total volume.
Ka = (H^+)(HCOO^-)/(HCOOH)
Solve for (H^+).
(HCOO^-) from above.
(HCOOH) from above.
Convert (H^+) to pH.
This is a buffered solution and you can use the Henderson-Hasselbalch equation if you wish. It is a little simpler because you may use moles/V and not actually determine the concns of each (since the volume cancels).
To calculate the pH of the resulting solution, we need to first consider the reaction that occurs when HCOOH (formic acid) and HCOONa (sodium formate) are mixed together.
HCOOH is a weak acid and HCOONa is its conjugate base. When they react, the HCOONa dissociates into its component ions (HCOO- and Na+), while HCOOH donates a proton (H+) to the solution.
The balanced chemical equation for the reaction is:
HCOOH + HCOONa -> HCOO- + HCOOH2+
Since HCOO- is a weak base, it will react with water to form OH- ions and HCOOH, resulting in a slight increase in OH- concentration and a corresponding decrease in H+ concentration.
To calculate the concentrations of HCOOH and HCOO- in the resulting solution, we can use the formula:
[HCOOH] = [initial HCOOH] - [HCOOH reacted]
[HCOO-] = [initial HCOONa] + [HCOOH reacted]
Given:
Initial volume of HCOOH solution (V1) = 50.0 mL
Initial concentration of HCOOH solution (C1) = 0.15 M
Initial volume of HCOONa solution (V2) = 75.0 mL
Initial concentration of HCOONa solution (C2) = 0.13 M
Ka of HCOOH = 1.8 x 10^(-4)
First, let's find the amount of reacted HCOOH using the balanced chemical equation and the concept of stoichiometry:
1 mole of HCOOH reacts with 1 mole of HCOONa, so the reaction is 1:1.
Using the formula:
[HCOOH reacted] = (C1 * V1) = (0.15 M) * (50.0 mL) = 7.5 mmol
[HCOOH reacted] = (C2 * V2) = (0.13 M) * (75.0 mL) = 9.75 mmol
Next, we can calculate the new concentrations of HCOOH and HCOO- in the resulting solution:
[HCOOH] = [initial HCOOH] - [HCOOH reacted]
[HCOOH] = (0.15 M) - (7.5 mmol / (V1 + V2))
[HCOOH] = (0.15 M) - (7.5 mmol / (50.0 mL + 75.0 mL))
[HCOO-] = [initial HCOONa] + [HCOOH reacted]
[HCOO-] = (0.13 M) + (9.75 mmol / (V1 + V2))
[HCOO-] = (0.13 M) + (9.75 mmol / (50.0 mL + 75.0 mL))
Now, we can calculate the concentration of H+ in the solution using the dissociation constant (Ka):
Ka = ([H+][HCOO-]) / [HCOOH]
[H+] = (Ka * [HCOOH]) / [HCOO-]
[H+] = (1.8 x 10^(-4) * [HCOOH]) / [HCOO-]
Finally, we can calculate the pH of the solution using the equation:
pH = -log[H+]
Substitute the calculated [H+] value into the equation to find the pH.