Answer to Question #344281 in Chemistry for soph

Question

13. If 23 grams of formic acid, HCOOH, are dissolved in 10L of water at 25oC, the [H+] is found to be 3.0x10-3 M. Determine the Ka for HCOOH. Hint: think ICE BOX. (6 marks)

Answer 1

Solution:

The molar mass of formic acid (HCOOH) is 46 g/mol

Therefore,

Moles of HCOOH = (23 g HCOOH) × (1 mol HCOOH / 46 g HCOOH) = 0.5 mol HCOOH

Molarity = Moles of solute / Liters of solution

Therefore,

Molarity of HCOOH solution = (0.5 mol) / (10 L) = 0.05 mol/L = 0.05 M

Formic acid (HCOOH) is a weak acid.

The balanced equation for the ionization of HCOOH is:

HCOOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HCOO⁻(aq)

or:

HCOOH(aq) ⇌ H⁺(aq) + HCOO⁻(aq)

Summarize the initial conditions, the changes, and the equilibrium conditions in the following ICE table:

[H⁺] = [HCOO⁻] = 3.0×10⁻³ M = 0.003 M

[HCOOH] = 0.047 M

Substitute the equilibrium concentrations into the expression for the acid ionization constant K_a:

Answer: K_a for HCOOH is 1.9×10⁻⁴