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Chapter 4: Problem 82

(a) How many milliliters of 0.120 \(\mathrm{M}\) HCl are needed to completely neutralize 50.0 \(\mathrm{mL}\) of 0.101 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of 0.125 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize 0.200 \(\mathrm{g}\) of NaOH? \((\mathrm{c})\) If 55.8 \(\mathrm{mL}\) of a BaCl \(_{2}\) solution is needed to precipitate all the sulfate ion in a 752 -mg sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what is the molarity of the BaCl\(_{2}\) solution?

Short Answer

Expert verified
In a short answer, (a) 84.2 mL of 0.120 M HCl are needed to completely neutralize 50.0 mL of 0.101 M Ba(OH)$_2$ solution. (b) 20.0 mL of 0.125 M H$_2$SO$_4$ are needed to neutralize 0.200 g of NaOH. (c) The molarity of the BaCl$_2$ solution is 0.095 M.

Step by step solution

01

Individual tasks

1. Calculate the number of moles 2. Convert the moles to volume (where applicable) 3. Proceed to the next problem
02

Problem a (part 1): Determine the Amount of Moles for HCl and \( \text{Ba(OH)}_2 \)

The balanced chemical equation will come out to be \[2\text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2\text{H}_2\text{O}\] In 50.0 mL of 0.101 M \(\text{Ba(OH)}_2\), the number of moles is: \(0.101 M \times 0.050 L = 0.00505 \text{ mole of Ba(OH)}_2\). Since the reaction with HCl is in a 2:1 ratio, the number of moles of HCl required will be \(2 \times 0.00505 = 0.0101 \text{ moles of HCl}\).
03

Problem a (part 2): Convert Moles to Volume for HCl

To convert this into volume, you need to use the molarity definition, M = moles/L. With reorganization you get V = moles / M, substituting the values: \(0.0101 \text{ moles} / 0.120 \text{M} = 0.0842L = 84.2 \text{mL of HCl}\).
04

Problem b (part 1): Determine Mole for \( \text{NaOH} \) and \( \text{H}_2\text{SO}_4 \)

The balanced equation for the neutralization of \( \text{NaOH} \) and \( \text{H}_2\text{SO}_4 \) is: \[2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}\] 0.200 g of \( \text{NaOH} \), which equates to \( \text{NaOH: 0.200 g} / 40.0 g\text{/mol} = 0.00500 \text{ moles of NaOH} \). The stoichiometry from the balanced equation tells us that 2 moles of \( \text{NaOH} \) reacts with 1 mole of \( \text{H}_2\text{SO}_4 \). So, we will need \( 0.00500 \text{ moles of NaOH} / 2 = 0.00250 \text{ moles of H}_2\text{SO}_4 \).
05

Problem b (part 2): Convert Moles to Volume for \( \text{H}_2\text{SO}_4 \)

The same as in part a, \( V = \text{moles} / \text{M} \). Substituting we find \( 0.00250 \text{ moles} / 0.125 \text{M} = 0.0200 \text{L} = 20.0 \text{mL of H}_2\text{SO}_4 \).
06

Problem c (part 1): Determine moles of \( \text{Na}_2\text{SO}_4 \) and \( \text{BaCl}_2 \)

The balanced equation for the reaction between the sulfate ion from \( \text{Na}_2\text{SO}_4 \) and \( \text{BaCl}_2 \) is: \[\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaCl}\] 752 mg of \( \text{Na}_2\text{SO}_4 \) would be equivalent to \( \text{Na}_2\text{SO}_4: 0.752 g / 142.04 g\text{/mol} = 0.00529 \text{ moles} \). The balanced equation suggests that 1 mole of \( \text{Na}_2\text{SO}_4 \) reacts with 1 mole of \( \text{BaCl}_2 \) to produce the precipitate. So, we will need the equivalent 0.00529 moles of \( \text{BaCl}_2 \).
07

Problem c (part 2): Determine molarity of \( \text{BaCl}_2 \)

Molarity (M) is given by the formula \( M = \text{moles} / V \). Using the given volume and the calculated moles of \( \text{BaCl}_2 \), we find, \( M = 0.00529 \text{ moles} / 0.0558 \text{L} = 0.095 \text{M} \) for the \( \text{BaCl}_2 \) solution.

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Most popular questions from this chapter

Label each of the following substances as an acid, base, salt, or none of the above. Indicate whether the substance exists in aqueous solution entirely in molecular form, entirely as ions, or as a mixture of molecules and ions. (a) HF, (b) acetonitrile,\(\mathrm{CH}_{3} \mathrm{CN},(\mathbf{c}) \mathrm{NaClO}_{4},(\mathbf{d}) \mathrm{Ba}(\mathrm{OH})_{2} .\)

A 4.36 -g sample of an unknown alkali metal hydroxide is dissolved in 100.0 \(\mathrm{mL}\) 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 theequivalence point has been reached, after 17.0 \(\mathrm{mL}\) 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 alkalimetal cation: \(\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+},\) or \(\mathrm{Cs}^{+} ?\)

Consider the following reagents: zinc, copper, mercury (density 13.6 \(\mathrm{g} / \mathrm{mL}\) , silver nitrate solution, nitric acid solution. (a) Given a 500 -mL Erlenmeyer flask and a balloon, can you combine two or more of the foregoing reagents to initiate a chemical reaction that will inflate the balloon? Write a balanced chemical equation to represent this process. What is the identity of the substance that inflates the balloon? (b) What is the theoretical yield of the substance that fills the balloon? (c) Can you combine two or more of the foregoing reagents to initiate a chemical reaction that will produce metallic silver? Write a balanced chemical equation to represent this process. What ions are left behind in solution? (d) What is the theoretical yield of silver?

Uranium hexafluoride, UF\(_{6}\), is processed to produce fuel for nuclear reactors and nuclear weapons. UF\(_{6}\) is made from the reaction of elemental uranium with \(\mathrm{ClF}_{3},\) which also produces \(\mathrm{Cl}_{2}\) as a by-product. (a) Write the balanced molecular equation for the conversion of U and \(\mathrm{ClF}_{3}\) into UF \(_{6}\) and \(\mathrm{Cl}_{2}\) . (b) Is this a metathesis reaction? (c) Is this a redox reaction?

Neurotransmitters are molecules that are released by nerve cells to other cells in our bodies, and are needed for muscle motion, thinking, feeling, and memory. Dopamine is a common neurotransmitter in the human brain. (a) Predict what kind of reaction dopamine is most likely to undergo in water: redox, acid-base, precipitation, or metathesis? Explain your reasoning. (b) Patients with Parkinson's disease suffer from a shortage of dopamine and may need to take it to reduce symptoms. An IV (intravenous fluid) bag is filled with a solution that contains 400.0 mg dopamine per 250.0 mL. of solution. What is the concentration of dopamine in the IV bag in units of molarity? (c) Experiments with rats show that if rats are dosed with 3.0 \(\mathrm{mg} / \mathrm{kg}\) of cocaine (that is, 3.0 mg cocaine per kg of animal mass), the concentration of dopamine in their brains increases by 0.75\(\mu M\) after 60 seconds. Calculate how many molecules of dopamine would be produced in a rat (average brain volume 5.00 \(\mathrm{mm}^{3} )\) after 60 seconds of a 3.0 \(\mathrm{mg} / \mathrm{kg}\) dose of cocaine.
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