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An \(8.65-g\) sample of an unknown group 2 metal hydroxide is dissolved in \(85.0 \mathrm{~mL}\) of water. An acid-base indicator is added and the resulting solution is titrated with \(2.50 \mathrm{M}\) \(\mathrm{HCl}(a q)\) solution. The indicator changes color, signaling that the equivalence point has been reached, after \(56.9 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the metal cation: \(\mathrm{Ca}^{2+}, \mathrm{Sr}^{2+}\), or \(\mathrm{Ba}^{2+}\) ?

Short Answer

Expert verified
The metal hydroxide is Sr(OH)₂ with a molar mass of 121.6 g/mol.

Step by step solution

01

Write the Balanced Reaction

The metal hydroxide can be expressed as \( M(OH)_2 \), where \( M \) is the group 2 metal. The balanced chemical equation for the reaction with HCl is: \[ M(OH)_2 + 2 HCl \rightarrow MCl_2 + 2 H_2O \] This indicates that one mole of \( M(OH)_2 \) reacts with two moles of HCl.
02

Calculate the Moles of HCl

We know the concentration of HCl is \( 2.50 \text{ M} \) and the volume of HCl used is \( 56.9 \text{ mL} \) or \( 0.0569 \text{ L} \). Use the formula \( \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \): \[ \text{Moles of HCl} = 2.50 \times 0.0569 = 0.14225 \text{ moles} \]
03

Calculate the Moles of Metal Hydroxide

From the balanced equation, we know that \( M(OH)_2 \) reacts with HCl in a 1:2 ratio. Therefore, the moles of \( M(OH)_2 \) is half the moles of HCl: \[ \text{Moles of } M(OH)_2 = \frac{0.14225}{2} = 0.071125 \text{ moles} \]
04

Calculate the Molar Mass of the Metal Hydroxide

We have \( 8.65 \text{ g} \) of \( M(OH)_2 \) and \( 0.071125 \) moles. Use the formula \( \text{Molar Mass} = \frac{\text{Mass}}{\text{Moles}} \): \[ \text{Molar Mass} = \frac{8.65}{0.071125} \approx 121.6 \text{ g/mol} \]
05

Determine the Identity of the Metal Cation

Compare the calculated molar mass \( 121.6 \text{ g/mol} \) with the known molar masses of group 2 metal hydroxides: \( \text{Ca(OH)}_2 \approx 74 \text{ g/mol} \), \( \text{Sr(OH)}_2 \approx 121.6 \text{ g/mol} \), \( \text{Ba(OH)}_2 \approx 171 \text{ g/mol} \). The calculated molar mass matches that of \( \text{Sr(OH)}_2 \).

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Key Concepts

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

Titration Analysis
Titration is a laboratory technique used to determine the concentration of a solute in a solution. In acid-base titration, a solution of known concentration (titrant) is added to a solution of unknown concentration until the reaction reaches the equivalence point. During this process, an indicator can be added to detect when the equivalence point, the point at which the amounts of acid and base are exactly balanced, is reached. The indicator changes color at this specific point, providing a visual signal.

In the given exercise, a titration analysis is performed to identify the molar mass of a group 2 metal hydroxide. The metal hydroxide is dissolved in water, and hydrochloric acid (HCl) is used as the titrant. By measuring the volume of HCl needed to reach the equivalence point, we can calculate the number of moles of HCl, and from that, determine the moles of the metal hydroxide present in the sample.
  • Volume of HCl used: 56.9 mL (0.0569 L)
  • Concentration of HCl: 2.50 M
  • Equivalence point signaled by color change
Chemical Reactions
Chemical reactions involve the transformation of substances through the breaking and forming of bonds. In the context of this exercise, the chemical reaction is between a group 2 metal hydroxide and hydrochloric acid.

The balanced chemical equation for the reaction involves one mole of metal hydroxide reacting with two moles of hydrochloric acid to produce one mole of a metal chloride and two moles of water:\[ M(OH)_2 + 2HCl \rightarrow MCl_2 + 2H_2O \]
This reaction highlights a key concept in stoichiometry, the calculation of reactants and products in chemical reactions. Understanding the mole ratio is crucial; here, the metal hydroxide reacts in a 1:2 ratio with HCl.These reactions are essential in titration as they allow us to deduce the molar relationships between reactants based on the volumes and concentrations used.
  • Balanced equation: 1:2 ratio of metal hydroxide to HCl
  • Produces metal chloride and water
  • Stoichiometry helps determine unknown sample characteristics
Group 2 Metal Hydroxides
The periodic table's group 2 metals form hydroxides that are integral in titration studies. Known as alkaline earth metals, they include: calcium (Ca), strontium (Sr), and barium (Ba). When these metals react with water, they produce metal hydroxides such as \( Ca(OH)_2 \), \( Sr(OH)_2 \), and \( Ba(OH)_2 \).

These hydroxides have characteristic molar masses which help in identifying the metal after experimental data are gathered. Knowing the molar mass can pinpoint the identity of the metal in the hydroxide:
  • \( Ca(OH)_2 \approx 74 \text{ g/mol} \)
  • \( Sr(OH)_2 \approx 121.6 \text{ g/mol} \)
  • \( Ba(OH)_2 \approx 171 \text{ g/mol} \)
In the exercise, the calculated molar mass of the unknown sample was approximately \( 121.6 \text{ g/mol} \), aligning with the molar mass of \( Sr(OH)_2 \), confirming that strontium is the metal present in the sample. Group 2 metal hydroxides play a significant role in understanding analytical chemistry principles and are fundamental to many industrial and scientific applications.

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Most popular questions from this chapter

You want to analyze a silver nitrate solution. (a) You could add \(\mathrm{HCl}(a q)\) to the solution to precipitate out \(\mathrm{AgCl}(s) .\) What volume of a \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) solution is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of a \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (b) You could add solid \(\mathrm{KCl}\) to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (c) Given that a \(0.150 \mathrm{M} \mathrm{HCl}(a q)\) solution costs \(\$ 39.95\) for \(500 \mathrm{~mL}\) and that KCl costs \(\$ 10 /\) ton, which analysis procedure is more cost-effective?

An aqueous solution contains \(1.2 \mathrm{~m} M\) of total ions. (a) If the solution is \(\mathrm{NaCl}(a q),\) what is the concentration of chloride ion? \((\mathbf{b})\) If the solution is \(\mathrm{FeCl}_{3}(a q),\) what is the concentration of chloride ion?

(a) A caesium hydroxide solution is prepared by dissolving \(3.20 \mathrm{~g}\) of \(\mathrm{CsOH}\) in water to make \(25.00 \mathrm{~mL}\) of solution. What is the molarity of this solution? (b) Then, the caesium hydroxide solution prepared in part (a) is used to titrate a hydroiodic acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between the caesium hydroxide and hydroiodic acid solutions. (c) If \(18.65 \mathrm{~mL}\) of the caesium hydroxide solution was needed to neutralize a \(42.3 \mathrm{~mL}\) aliquot of the hydroiodic acid solution, what is the concentration (molarity) of the acid?

The concept of chemical equilibrium is very important. Which one of the following statements is the most correct way to think about equilibrium? (a) If a system is at equilibrium, nothing is happening. (b) If a system is at equilibrium, the rate of the forward reaction is equal to the rate of the back reaction. (c) If a system is at equilibrium, the product concentration is changing over time. [Section 4.1]

The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, \(\mathrm{AsO}_{4}^{3-},\) what mass of sodium arsenate would be present in a \(1.00-\mathrm{L}\) sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as $$ \mathrm{ppb}=\frac{\mathrm{g} \text { solute }}{\mathrm{g} \text { solution }} \times 10^{9} $$

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