Question

to 21. In the laboratory, you are asked to recommends using oxalic acid (K = 5.374 your laboratory partner? you are asked to prepare a buffer at pH 5. Your laboratory partner alic acid (K. = 5.37 x 10") and it salt. What do you recommend to 22. Formic acid (HCOOH) is a monoprotic acid. Is formic acid a weak acid or strong acid? How do you know The K. value of formic acid is 1.78 x 10. What is the pk, value of formic acid? Suppose you have a 0.15M aqueous solution of formic acid. Write the chemical equation and acid dissociation constant expression d) Calculate the pH of the solution at equilibrium. b) The K What is you know? aqueous and acid disso Formic aad is a strong aade In the formule, the natio for both are the same, so if there is more than H, it's strong. But, it is usually first in the formula, it's a strong acid. b) in (1.78*10-4)=-8,6% C) HCOOH —> COOK+X & HOOH ? Alor tint 015110 120 C-X + x + x of E 115x1 x 11 x 1.78*10-4 = [x] [x] ./5-x 15-178*10-4 JCBOBENKO FALL 2019 CHEM 341 x=267x105 los 5.1740's 7a 517103 2.29

Answer 1

21) The pKa of oxalic acid is calculated:

pKa = - log 5.37x10 ^ -2 = 1.27

The pKa of oxalic acid is very low for the required pH, it is proposed to use an acid with a pKa closer to:

pKa = 10 ^ -5

22) a) It is a weak acid, since it dissociates only partially in aqueous solution, it has a Ka that relates its balance.

b) The pKa is calculated:

pKa = - log 1.78x10 ^ -4 = 3.75

c) You have the reaction:

HCOOH + H2O = HCOO- + H3O +

d) There is the expression of Ka:

Ka = [HCOO-] * [H3O +] / [HCOOH]

1.78x10 ^ -4 = X ^ 2 / 0.15

It clears X = 0.0052 M

The pH is calculated:

pH = - log 0.0052 = 2.29

17) The enthalpy change is the heat transferred at constant pressure.

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