Question

What are the concentrations of \(\mathrm{HSO}_{4}^{-}, \mathrm{SO}_{4}^{2-},\) and \(\mathrm{H}^{+}\) in a \(0.20 \mathrm{M} \mathrm{KHSO}_{4}\) solution? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a strong acid: \(K_{\mathrm{a}}\) for \(\left.\mathrm{HSO}_{4}^{-}=1.3 \times 10^{-2} .\right)\)

Short answer

The concentrations of HSO4-, SO42-, and H+ are 0.187 M, 0.013 M, and 0.013 M, respectively.

Step-by-step solution

Identify the Reaction and Write the Equilibrium Expression

First, write the equation for the dissociation of HSO4- in water: \nHSO4- (aq) ⇌ H+ (aq) + SO42- (aq).\nThe equilibrium expressions for the reaction can be written as: \[ K_{a} = \frac{[H^{+}][SO_{4}^{2-}]}{[HSO_{4}^{-}]} \]

Initial Concentration of HSO4-

The problem states that we start with a 0.20M solution of KHSO4. KHSO4 is a salt that fully ionizes in solution to give K+ and HSO4-. Since it fully ionizes, the initial concentration of HSO4- is also 0.20 M.

Calculate the Concentration of H+ and SO42-

Since HSO4- ionizes to produce H+ and SO42-, we can assume that the concentration of H+ is equal to the concentration of SO42- since one of each is produced for every HSO4- ion that ionizes. To find these concentrations, we can set up an equation based on the Ka expression. \nLet x be the concentration of H+ (which is also the concentration of SO42-). Therefore, Ka = (x)(x) / (0.20M - x). Solving this quadratic equation, we find that x = 0.013 M. So, the concentration of H+ and SO42- is both 0.013 M.

Calculate the remaining concentration of HSO4-

The initial concentration of HSO4- was 0.20 M, and x amount of it ionized, so the remaining concentration of HSO4- is simply the initial concentration minus the amount that ionized: 0.20 M - 0.013 M = 0.187 M

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is like learning a delicate dance, where reactants and products are the dancers, and the dance floor is the chemical reaction. In solutions, some reactions reach a point where the reactants and products are present in fixed proportions that don't change over time. This state is known as chemical equilibrium. At equilibrium, the rate of the forward reaction, where reactants turn into products, is equal to the rate of the reverse reaction, where products revert to reactants.
For instance, when the compound potassium hydrogen sulfate (KHSO₄) dissolves in water, it forms hydronium ions (H⁺) and sulfate ions (SO₄^2-). Initially, the ions are produced at a high rate, but as the reaction proceeds, a reverse process begins where the ions combine back to form KHSO₄. Eventually, the rate of these two opposing processes becomes equal, and the concentrations of the ions become constant – a perfect illustration of chemical equilibrium.

Acid Dissociation Constant

Let's explore the world of split personalities within chemistry — specifically, an acid's dissociation in a solution. The acid dissociation constant (K_a) is a numerical value that measures the strength of an acid in solution. In technical terms, it quantifies the extent to which an acid can donate a proton (H+) to the solvent, typically water. The larger the value of K_a, the stronger the acid is, as it means a greater fraction of acid molecules have donated their protons to become ions.
For HSO₄^–, this constant helps us understand exactly how many of its molecules will separate into ions. With a K_a of 1.3 × 10^–2, it indicates a relatively strong ability to ionize – it's eager to release its H⁺ and form SO₄^2-, turning the solution acidic.
Our given exercise takes advantage of this understanding to predict the concentrations of ions in the solution at equilibrium. Knowing the K_a allows us to set up our equilibrium expression and solve for the mysterious 'x' – the concentration of our dissociated ions.

Concentration Calculations

Bon voyage! We're embarking on a quest to determine how much of a substance exists within a solution — and this quest involves concentration calculations. The unit moles per liter (M), or molarity, is employed in these calculations to express concentration. When a compound like KHSO₄ is dissolved in water, its disintegration into ions is our chief concern.
In our example, we began with a 0.20 M solution of KHSO₄. This concentration tells us the initial amount of HSO₄^– available to dissociate. The tale begins with all characters in place, but as the story of dissociation unfolds, some HSO₄^– characters leave their original form to become the ions H⁺ and S0₄^2-, diminishing their initial population. By writing out and solving a mathematical equation, we uncover the final concentrations of all players: HSO₄^– stays at a concentration of 0.187 M, while the new characters, H⁺ and S0₄^2-, each emerge at a concentration of 0.013 M.
The art of concentration calculations lies in understanding chemical behavior and employing algebra to reveal the various concentrations at equilibrium, guiding us through the fog of a complex chemical landscape.