Question

A solution is prepared by mixing \(45.00 \mathrm{~mL}\) of \(0.022 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) with \(13.00 \mathrm{~mL}\) of \(0.0014 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3} .\) Assume that volumes are additive. (a) Will precipitation occur? (b) Calculate \(\left[\mathrm{Ag}^{+}\right],\left[\mathrm{CO}_{3}^{2-}\right],\left[\mathrm{Na}^{+}\right],\) and \(\left[\mathrm{NO}_{3}^{-}\right]\) after equilibrium is established.

Short answer

Answer: The final concentrations of ions in the solution are: [Ag+] = 0.0127 M [CO3^2-] ≈ 0 M [Na+] = 0.000746 M [NO3-] = 0.0203 M

Step-by-step solution

Calculate initial concentrations of ions before mixing

We need to find the initial concentrations of Ag+, CO3^2-, Na+, and NO3- ions. To do this, we will use the equation: Concentration = (Molarity × Volume) / Total volume For AgNO3 solution, we have: Initial concentration of Ag+ ions = (0.022 M × 45.00 mL) / (45.00 mL + 13.00 mL) = 0.0167 M Initial concentration of NO3- ions = 0.0167 M (equal to the concentration of Ag+) For the Na2CO3 solution, we have: Initial concentration of CO3^2- ions = (0.0014 M × 13.00 mL) / (45.00 mL + 13.00 mL) = 0.000373 M Initial concentration of Na+ ions = 2 × CO3^2- concentration = 0.000746 M

Determine if precipitation will occur

To determine if a precipitate will form, we need to compare the reaction quotient (Q) with the solubility product constant (Ksp) for Ag2CO3. If Q > Ksp, precipitation will occur. The Ksp of Ag2CO3 is 8.1 × 10^(-12). The reaction quotient Q can be calculated as follows: Q = [Ag+]^2 × [CO3^2-] Substitute the values: Q = (0.0167)^2 × (0.000373) = 1.04 × 10^(-7) Now compare Q with Ksp: 1.04 × 10^(-7) > 8.1 × 10^(-12) Since Q > Ksp, precipitation will occur.

Calculate concentrations of ions after equilibrium

To calculate the concentrations of ions after equilibrium, we need to perform the following steps: a. Write the balanced equation: AgNO3 + Na2CO3 ⇒ Ag2CO3 (s) + 2 NaNO3 b. Calculate the moles of the limiting reactant using stoichiometry: Moles of AgNO3 = (0.022 M) × (45.00 mL / 1000) = 9.9 × 10^(-4) mol Moles of Na2CO3 = (0.0014 M) × (13.00 mL / 1000) = 1.82 × 10^(-5) mol Na2CO3 is the limiting reactant, so we can then calculate the amount of Ag2CO3 formed: Moles of Ag2CO3 = 0.5 × moles of Na2CO3 = 9.1 × 10^(-6) mol c. Subtract the moles of the limiting reactant from the initial moles. Calculate the final concentrations after equilibrium: - Ag+ concentration = (0.022 M) - (0.002 moles Ag+ / total volume) = 0.0127 M - CO3^2- concentration = (0.0014 M) - (0.0000182 moles CO3^2- / total volume) ≈ 0 M (negligible) - Na+ concentration = (0.000373 M) + (0.000363 moles Na+ / total volume) = 0.000746 M - NO3- concentration = (0.0167 M) + (0.000363 moles NO3- / total volume) = 0.0203 M The final concentrations after equilibrium are: [Ag+] = 0.0127 M [CO3^2-] ≈ 0 M [Na+] = 0.000746 M [NO3-] = 0.0203 M