A solution is prepared by mixing 35.00 mL of a 0.061 M...

A solution is prepared by mixing 35.00 mL of a 0.061 M solution of zinc nitrate with 20.0 mL of KOH with a pH of 9.00 . Assume that volumes are additive. (a) Will precipitation occur? (b) Calculate $\left[\mathrm{Zn}^{2+}\right],\left[\mathrm{NO}_3^{-}\right],\left[\mathrm{K}^{+}\right]$, and the pH after equilibrium is established.

A solution is prepared by mixing 35.00 mL of a 0.061 M solution of zinc nitrate with 20.0 mL of KOH with a pH of 9.00 . Assume that volumes are additive.
(a) Will precipitation occur?
(b) Calculate $\left[\mathrm{Zn}^{2+}\right],\left[\mathrm{NO}_3^{-}\right],\left[\mathrm{K}^{+}\right]$, and the pH after equilibrium is established.

Key Concepts

pH and pOH Calculations

pH and pOH calculations are central in evaluating the acidity or basicity of a solution. They involve converting the concentration of hydrogen ions or hydroxide ions into the pH scale, which is particularly useful in analyzing equilibrium conditions in mixed solutions.

Dilution and Mixing of Solutions

When solutions are combined, their solute concentrations change due to the increase in total volume. Dilution calculations are used to determine the new concentrations of ions by accounting for the initial moles of solute and the final, mixed volume.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative analysis of reactants and products in a chemical reaction. It enables the determination of how much of each species participates in the reaction, which is essential for predicting outcomes like precipitation and for calculating concentrations after the reaction.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. It is used to predict whether a precipitate will form by comparing the ionic product of the ions in solution (Q) to the Ksp value; if Q exceeds Ksp, precipitation occurs.

Precipitation Reactions

Precipitation reactions occur when two soluble salts are mixed and one of the resulting products is insoluble, forming a solid precipitate. This concept is important in understanding how mixing solutions can lead to the formation of a new phase as certain ionic combinations exceed solubility limits.

Transcript

00:01 This question contains multiple parts and requires you to review the concept of limiting reactant.

00:11 We have two reactants, for both of which we are given quantities.

00:18 Both of these quantities will decrease.

00:21 One will decrease to zero.

00:23 That will be the limiting reactant.

00:26 The other one will decrease not to zero, and some will be left over.

00:31 The maximum amount of product that can be made will be produced when the limiting reactant is completely consumed.

00:42 To identify the limiting reactant, you need to calculate the amount of product that can be created by each reactant, and the one that produces the smallest amount of product will be the limiting reactant.

00:56 So part a asks you to calculate the moles of precipitate that will form in the reaction, to do this, you'll calculate the moles of precipitate that will form for each of the reactants if each were consumed completely, and the one that produces the smallest amount will be the limiting reactant.

01:18 Before we can do that, we must first identify the chemical reaction that is occurring.

01:24 We have zinc nitrate reacting with potassium hydroxide.

01:29 If zinc nitrate reacts with potassium hydroxide and the cationons switch places, we'll get potassium nitrate, which according to the solubility rules, is soluble, and zinc hydroxide, which according to the solubility rules, is insoluble.

01:46 So to simplify things, the net ionic equation is zinc 2 plus reacting with two hydroxides, producing the insoluble precipitate zinc hydroxide.

02:00 Everything else are spectator ions.

02:03 Let's now calculate the mass of the.

02:07 Precipitate zinc hydroxide that can be created if all of the zinc nitrate were consumed all 200 milliliters we'll convert the 200 milliliters into liters and then convert those liters into moles of zinc 2 plus or zinc nitrate because the stoichi are the relationship is one to one by multiplying by the molar mass when we know the moles of zinc 2 plus, we can use the stoichiometry up here to calculate the moles of zinc hydroxide, which can be created.

02:46 It's a one -to -one relationship also.

02:49 And we get 0 .2 -00 moles zinc hydroxide if all of the zinc nitrate were consumed...

Please give Ace some feedback