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A solution is made by mixing 15.0 \(\mathrm{g}\) of \(\mathrm{Sr}(\mathrm{OH})_{2}\) and 55.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HNO}_{3}\) . (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resulting solution acidic or basic?

Short Answer

Expert verified
(a) The balanced equation for the reaction between Sr(OH)鈧 and HNO鈧 is: \( Sr(OH)_2 + 2HNO_3 \rightarrow Sr(NO_3)_2 + 2H_2O \). (b) The concentrations of ions remaining in the solution are: Sr虏鈦 ions: 2.04 M NO鈧冣伝 ions: 0.40 M OH鈦 ions: 4.08 M (c) The resulting solution is basic as the OH鈦 ion concentration is greater than the H鈦 ion concentration.

Step by step solution

01

(a) Balanced equation

To write a balanced equation for the reaction between Strontium hydroxide (Sr(OH)2) and Nitric acid (HNO3), we first identify that they react with each other as an acid-base reaction, forming a salt (Strontium nitrate) and water. The balanced equation will be: \( Sr(OH)_2 + 2HNO_3 \rightarrow Sr(NO_3)_2 + 2H_2O \)
02

(b1) Moles and concentration of the reactants

First, we will determine the moles and concentrations of Sr(OH)2 and HNO3 in the initial solution. Given: Mass of Sr(OH)2 = 15.0 g Volume of HNO3 solution = 55.0 mL Concentration of HNO3 solution = 0.200 M Molecular weight of Sr(OH)2 = 121.63 g/mol (38.0 g/mol (Sr) + 2(16.0 g/mol (O) + 1.0 g/mol (H))) Now, calculate the moles of Sr(OH)2: Moles of Sr(OH)2 = \( \frac{15.0 \,\text{g}}{121.63 \,\text{g/mol}} = 0.1233 \,\text{mol} \) Calculate the moles of HNO3: Moles of HNO3 = 55.0 mL 脳 0.200 mol/L = 0.0110 mol
03

(b2) Determine the limiting reactant

To find the limiting reactant, we can compare the stoichiometric ratio of the reactants with the moles present in the solution. From the balanced equation, the ratio of Sr(OH)2 to HNO3 is 1:2. Compare the moles against this ratio: \( \frac{\text{moles of Sr(OH)}_2}{\text{moles of HNO}_3} = \frac{0.1233 \,\text{mol}}{0.0110 \,\text{mol}} = 11.21 \) Since 11.21 > 1, Sr(OH)2 is in excess and HNO3 is the limiting reactant.
04

(b3) Calculate the moles of ions remaining

Based on the limiting reactant and the balanced equation, calculate the moles of ions remaining in the solution after the reaction is complete. For Sr(OH)2: Moles of Sr(OH)2 remaining = \(0.1233 \,\text{mol} - 0.0110 \,\text{mol} = 0.1123 \,\text{mol} \) For NO3-: Moles of NO3- produced: \( 2 \times 0.0110 \,\text{mol} = 0.0220 \,\text{mol} \)
05

(b4) Calculate the concentrations of ions remaining

To calculate the concentration (M) of each ion remaining, divide the moles by the total volume of the solution. The total volume is given as the original volume of HNO3, which is 55.0 mL. Concentration of Sr2+ ions = \( \frac{0.1123 \,\text{mol}}{0.0550 \,\text{L}} = 2.04 \,\text{M} \) Concentration of NO3- ions = \( \frac{0.0220 \,\text{mol}}{0.0550 \,\text{L}} = 0.40 \,\text{M} \) Concentration of OH- ions = \( 2 脳 \frac{0.1123 \,\text{mol}}{0.0550 \,\text{L}} = 4.08 \,\text{M} \)
06

(c) Determine if the solution is acidic or basic

A solution is acidic if the concentration of H+ ions is greater than the concentration of OH- ions, and basic if the concentration of OH- ions is greater than the concentration of H+ ions. In this case, the OH- ion concentration is 4.08 M, and there are no additional H+ ions in the solution (since all the HNO3 was reacted). Thus, the solution is basic.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is at the heart of chemistry, dealing with the quantitative relationships between reactants and products in a chemical reaction. By understanding stoichiometry, one can predict how much product will form from a given amount of reactants, which is essential in chemical manufacturing and laboratory experiments.

In the context of an acid-base titration calculation, stoichiometry allows us to calculate the exact amount of an acid required to neutralize a base, or vice versa. The balanced chemical equation provides a ratio of the reactants, which is used to calculate the amount of products formed or the amount of reactant needed for complete reaction. In our exercise, the balanced equation \( Sr(OH)_2 + 2HNO_3 \rightarrow Sr(NO_3)_2 + 2H_2O \) indicates that one mole of strontium hydroxide reacts with two moles of nitric acid to produce one mole of strontium nitrate and two moles of water. This stoichiometric ratio is pivotal for all subsequent calculations.
Limiting Reactant Determination
The limiting reactant in a chemical reaction is the substance that is completely consumed first, thus determining the endpoint of the reaction. Identifying the limiting reactant is crucial because it allows us to calculate the theoretical yield of the products.

In the exercise, we compared the actual mole ratio of the reactants to the stoichiometric mole ratio from the balanced equation. With a greater actual mole ratio of \( Sr(OH)_2 \) to \( HNO_3 \) than required, we infer that \( HNO_3 \) is the limiting reactant. After it is used up, the reaction stops, and any remaining \( Sr(OH)_2 \) is considered excess. This concept is integral to the computation of the concentration of ions remaining in the solution because it allows us to direct our focus solely towards the amounts consumed and produced by the limiting reactant.
Molarity Calculation
Molarity, symbolized as 'M', is a measure of concentration that represents the number of moles of a solute per liter of solution. This concept is a staple in solution chemistry, because it allows chemists to communicate how concentrated or dilute a solution is.

To determine the molarity of each ion remaining in the solution, we simply divide the moles of each ion by the total volume of the solution in liters. For example, in the solution after the reaction, the concentration of \( Sr^{2+} \) ions was found by dividing the moles of \( Sr^{2+} \) remaining by 0.0550 L (the volume of the solution), yielding a concentration of 2.04 M. This calculation is performed for all species remaining in the solution to determine their final concentrations.
Acid-Base Neutralization
Acid-base neutralization is a type of chemical reaction in which an acid and a base react to form water and a salt. This reaction is fundamental in various fields, including industrial processes, biological systems, and environmental science. The neutralization typically results in a change in the pH of the solution, which can be used to determine the acid or basic nature of the resulting solution.

In our exercise, because \( HNO_3 \) is the limiting reactant and all of it reacts, there are no remaining \( H^+ \) ions in the solution. The solution contains excess \( OH^- \) ions, resulting in a basic solution. This is consistent with the definition of a basic solution being one where the concentration of \( OH^- \) ions exceeds that of \( H^+ \) ions. Understanding this principle supports the response to whether the solution is acidic or basic after the reaction has occurred.

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Most popular questions from this chapter

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: \begin{equation} \begin{array}{r}{2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+} \quad\\\ {2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g)}\end{array} \end{equation} Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If 27 \(\mathrm{mL}\) of 6.0 \(\mathrm{MH}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

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Determine the oxidation number of sulfur in each of the following substances: (a) barium sulfate, \(\mathrm{BaSO}_{4}\) (b) sulfurous acid, \(\mathrm{H}_{2} \mathrm{SO}_{3},(\mathbf{c})\) strontium sulfide, \(\mathrm{SrS},(\mathbf{d})\) hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{S} .\) (e) Locate sulfur in the periodic table in Exercise 4.47; what region is it in? (f) Which region(s) of the periodic table contains elements that can adopt both positive and negative oxidation numbers?

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