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)
3 answers
Write the equation. Convert to moles KOH and moles formic acid, determine which reagent is in excess, calculate the amount of the excess reagent plus the amount of salt formed, then use the Henderson-Hasselbalch equation.
HCOOH + OH- --> COOH- + H20
Kb = [OH-][HCOOH]/[COOH-] = x^2/(0.0714-x)
= 5.55 x 10^-11
x=1.99 x 10^-6
pOH = 5.70
pH = 14 - 5.70 = 8.30
^ Will that also work?
Kb = [OH-][HCOOH]/[COOH-] = x^2/(0.0714-x)
= 5.55 x 10^-11
x=1.99 x 10^-6
pOH = 5.70
pH = 14 - 5.70 = 8.30
^ Will that also work?
Yes it will work and what I wrote for you will not work BECAUSE I didn't work the problem and I just assumed that the acid would be in excess and that would give you a formate/formic acid buffer. Actually, it gives you an EXACT neutralization where neither KOH NOR formic acid are in excess and you have only the salt present (and no buffer so the H-H equation won't work). Therefore, it must be done by the hydrolysis of the salt and you have approached it correctly. I see one small correct; in this step you wrote,
Kb = [OH-][HCOOH]/[COOH-] = x^2/(0.0714-x) . The denominator should be(COOK) [or (COO^-)] which is the concn of the salt. You substituted the correct value for it in your solution.
Kb = [OH-][HCOOH]/[COOH-] = x^2/(0.0714-x) . The denominator should be(COOK) [or (COO^-)] which is the concn of the salt. You substituted the correct value for it in your solution.
1 answer