Question

\(\begin{array}{ll}\text { The } \text { equilibrium } & \text { carbonate } \text { ion }\end{array}\) concentration after equal volumes of \(0.7 \mathrm{M}-\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(0.7 \mathrm{M}-\mathrm{HCl}\) solutions are mixed, is \(\left(K_{\mathrm{al}}\right.\) and \(K_{\mathrm{a} 2}\) for \(\mathrm{H}_{2} \mathrm{CO}_{3}\) are \(4.9 \times 10^{-6}\) and \(4.0 \times 10^{-11}\), respectively) (a) \(0.7 \mathrm{M}\) (b) \(0.35 \mathrm{M}\) (c) \(0.002 \mathrm{M}\) (d) \(0.001 \mathrm{M}\)

Short answer

The concentration of carbonate ions after reaction will be approximately equal to the second dissociation constant, Ka2, which is given as 4.0 x 10^-11 M, so the answer is (d) 0.001 M.

Step-by-step solution

Determine the Balanced Chemical Reaction

When solutions of Na2CO3 and HCl are mixed, they react in a 1:2 stoichiometry, with each carbonate ion reacting with two H+ ions, forming H2CO3 (carbonic acid). The balanced chemical equation for the reaction is: Na2CO3 + 2HCl → 2NaCl + H2CO3.

Calculate Initial Moles of Reactants

Since equal volumes of 0.7 M Na2CO3 and 0.7 M HCl are mixed, the initial moles of HCl will be double that of carbonate ions (because of the 1:2 stoichiometry). However, because they're of the same concentration and equal volume, carbonate ions are completely neutralized, and moles of H2CO3 formed will be the same as moles of carbonate ions initially present.

Determine the Reaction of H2CO3 in Water

H2CO3 is a weak acid and partly dissociates in water to form HCO3- (bicarbonate) and H+ in its first dissociation step, and CO3^2- (carbonate) and another H+ in the second dissociation step. We need to consider these equilibria to determine the final carbonate concentration.

Apply the Acid Dissociation Equilibrium

Write the expressions for the acid dissociation constants (Ka1 and Ka2) for H2CO3. Since Ka1 >> Ka2, the first dissociation will nearly proceed to completion but the second will not, thus we can consider only the first dissociation for approximate concentration calculation. Use the formula Ka1 = [HCO3-] x [H+] / [H2CO3] to calculate [HCO3-], assuming initial [H2CO3] is approximately the final [HCO3-] due to strong first dissociation.

Find [CO3^2-] Using the Second Dissociation

The second dissociation of H2CO3 is not complete, so we use Ka2 to calculate the concentration of carbonate ions. The formula is Ka2 = [CO3^2-] x [H+] / [HCO3-]. Since the initial H2CO3 and HCO3- are roughly in equal concentration, we simplify the equation to Ka2 = [CO3^2-]x[H+], and since [H+] is very small, we can approximate [CO3^2-] to be equal to Ka2.

Compare with Given Options and Conclude

The carbonate concentration calculated will be very small, closely approximating the given value of Ka2, because of the negligible second dissociation. Therefore, we compare the calculated [CO3^2-] with the provided options and choose the closest one.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is similar to watching a tug of war where neither team is strong enough to outdo the other; the forces are balanced, and no apparent change occurs as time passes. In chemistry, this battle happens at the molecular level within a closed system where the rate of the forward reaction equals the rate of the reverse reaction. The concentrations of reactants and products remain constant, not because the reactions have stopped, but because they proceed at equal rates in opposite directions.

For instance, in a typical acid-base neutralization that forms a weak acid, the products can partially revert to reactants until a stable ratio is reached, which is pointed out by an equilibrium constant. This immutable ratio is critical because it allows chemists to predict the concentrations of all species in a reaction mixture at equilibrium.

Acid-Base Reactions

Imagine acid-base reactions as a dance between protons and chemical species in solution. When an acid (proton donor) and a base (proton acceptor) meet in aqueous solution, they 'dance' together, exchanging protons in a process known as neutralization. This invites new products to the dance floor: a salt and usually water. The acid's tendency to lose a proton and the base's affinity to gain one are critical in predicting the course of such reactions. When weak acids or bases are involved, they don't give up or take their dancing partners (protons) so readily. This reluctance gives rise to incomplete reactions and establishes a dynamic equilibrium, significantly affecting the equilibrium carbonate concentration in our discussed scenario.

Stoichiometry

Stoichiometry can be likened to a recipe in cooking—it tells you exactly how much of each ingredient you need to end up with the desired amount of product. In chemical reactions, stoichiometry is built on the conservation of mass law, dictating that atoms won’t disappear or appear from nowhere; they simply rearrange to form new substances.

By using the balanced chemical equation, we can calculate precisely how much of each reactant is needed to produce a certain amount of product. For instance, the stoichiometry of a reaction between a carbonate and an acid like HCl is vital in understanding that it takes two acid molecules to neutralize one carbonate molecule, which becomes the foundation for determining the equilibrium concentrations after mixing the reacting solutions.

Acid Dissociation Constants

An acid dissociation constant, denoted as Ka, measures the strength of an acid in solution. It is the equilibrium constant for the dissociation reaction of the acid into its ions. Each acid has a particular Ka value, which provides a glimpse into its personality; a high Ka value reveals a strong acid, while a low value means a weaker player in the acid game. Metaphorically thinking, Ka tells us how eager an acid is to split and let loose its protons into the dance of the solution.

For our exercise on carbonate equilibrium, the two stages of acid dissociation (Ka1 and Ka2) shed light on the tendency of carbonic acid to part with its protons. Ka1 is much larger than Ka2, indicating that the first proton is lost readily, contributing significantly to the bicarbonate concentration. However, the loss of the second proton, dictated by Ka2, is much more begrudgingly done, greatly limiting the carbonate ion concentration in the end.