How many grams of potassium formate, KHCOO, must be added to 500. mL of a 0.0700

Question

How many grams of potassium formate, KHCOO, must be added to 500. mL of a 0.0700 M solution of formic acid, HCOOH, to produce a buffer solution with a pH of 3.50? (Any change in the volume of the solution due to the addition of solid potassium formate is insignificant.) K_a = 1.8 times 10^-4 for HCOOH. a. 1.0 g b. 5.2 g c. 6.7 g d. 1.7 g Calculate the solubility product constant for aluminum hydroxide. Its molar solubility is 2.9 times 10^-9 mole per liter at 25 degree C. a. 9.8 times 10^-26 b. 4.9 times 10^-26 c. 7.1 times 10^-35 d. 2.1 times 10^-34 e. 1.9 times 10^-33

Explanation / Answer

Q55.

this is a buffer so

pH = pKa + log(A-/HA)

3.50 = 3.75 + log(A-/HA)

3.50 = 3.75 + log(A-/(0.5*0.07))

mol of formate = (0.5*0.07)*10^(3.50-3.75) = 0.019681

mass = mol*MW = 0.019681*84.12 = 1.65556 g required

choose 1.7 g , which is D

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