# Question Video: Calculating the Value of K_𝑎 for a Solution of Propanoic Acid Chemistry

Calculate the K_𝑎 value of a 0.2 M aqueous solution of propanoic acid with a concentration of H⁺ ions of 1.62 × 10⁻³ M. Give your answer to 1 decimal place. [A] 4.8 × 10⁻⁴ mol/L [B] 6.6 × 10⁻⁵ mol/L [C] 1.6 × 10⁻² mol/L [D] 5.2 × 10⁻⁷ mol/L [E] 1.3 × 10⁻⁵ mol/L

04:03

### Video Transcript

Calculate the K𝑎 value of a 0.2-molar aqueous solution of propanoic acid with a concentration of H+ ions of 1.62 times 10 to the negative three molar. Give your answer to one decimal place. (A) 4.8 times 10 to the negative four moles per liter. (B) 6.6 times 10 to the negative five moles per liter. (C) 1.6 times 10 to the negative two moles per liter. (D) 5.2 times 10 to the negative seven moles per liter. Or (E) 1.3 times 10 to the negative five moles per liter.

Propanoic acid is CH3CH2COOH. We know this from the name stem “prop,” meaning three carbons in the chain, “an” telling us there are single bonds between carbon atoms, and the suffix -oic acid telling us this is a carboxylic acid with a carboxyl group. Propanoic acid is a weak acid, meaning in solution it dissociates to a small degree, forming an equilibrium according to this equation. The dissociation products are the ions CH3CH2COO− and the hydrogen ion. We are asked to find K𝑎, the acid dissociation constant for this weak acid.

We are told the starting concentration of the acid, 0.2 molar. And we are told the hydrogen ion concentration in solution, 1.62 times 10 to the negative three molar. Let’s start by writing the expression for K𝑎. K𝑎 is equal to the molar concentrations of the ion products, which are the propanoate ion and the hydrogen ion, multiplied with each other divided by the molar concentration of propanoic acid, the reactant. Square brackets refer to molar concentration, or molarity, which is moles per liter or moles per decimeter cubed.

We are given the denominator value in terms of molarity, 0.2 molar. And we are given the hydrogen ion concentration in terms of molarity too, 1.62 times 10 to the negative three molar. But we don’t know the concentration of the propanoate ion. However, from the balanced equation, we know that the molar ratio of the anion to the hydrogen cation is one as to one. And so the concentration of the propanoate anion must be the same as that of the hydrogen ion.

Let’s now put in our values. We get 1.62 times 10 to the negative three molar for the concentration of the anions multiplied by 1.62 times 10 to the negative three molar for the concentration of the hydrogen ions divided by 0.2 molar, which is the concentration of propanoic acid. And the value of the answer is 1.31 times 10 to the negative five. These two units can cancel, and so the unit in the answer is molar, which is moles per liter. Note that sometimes K𝑎 and its relative K𝑏, the base dissociation constant, are expressed without units.

We were asked to give our answer to one decimal place. So let’s round off, and we get 1.3 times 10 to the negative five moles per liter, which corresponds with answer option (E). The K𝑎 value for this propanoic acid equilibrium is 1.3 times 10 to the negative five moles per liter.