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Chapter 4: Problem 39

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Short Answer

Expert verified
The short answers for each part of the exercise are as follows: (a) The net ionic equation is: \(2H^{+}(aq) + 2OH^{-}(aq) \rightarrow 2H_{2}O(l)\) (b) The net ionic equation is: \(Cu(OH)_{2}(s) + 2H^{+}(aq) \rightarrow Cu^{2+}(aq) + 2H_{2}O(l)\) (c) The net ionic equation is: \(Al(OH)_{3}(s) + 3H^{+}(aq) \rightarrow Al^{3+}(aq) + 3H_{2}O(l)\)

Step by step solution

01

Predict the products

We have an acid-base reaction (double displacement). To predict the products, we swap the cations of the reactants: Ca虏鈦 and 2OH鈦 react with 2H鈦 and Br鈦 to form CaBr鈧 and H鈧侽.
02

Balance the molecular equation

The balanced molecular equation is: 2HBr(aq) + Ca(OH)鈧(aq) 鈫 CaBr鈧(aq) + 2H鈧侽(l)
03

Write the complete ionic equation

Breaking each compound into its individual ions, we get: 2H鈦(aq) + 2Br鈦(aq) + Ca虏鈦(aq) + 2OH鈦(aq) 鈫 Ca虏鈦(aq) + 2Br鈦(aq) + 2H鈧侽(l)
04

Write the net ionic equation

Cancel the spectator ions (Ca虏鈦 and 2Br鈦) and obtain the net ionic equation: 2H鈦(aq) + 2OH鈦(aq) 鈫 2H鈧侽(l) (b) Cu(OH)鈧(s) + HClO鈧(aq) 鈫
05

Predict the products

We have an acid-base reaction with Cu虏鈦 and 2OH鈦 reacting with H鈦 and ClO鈧勨伝 to form Cu(ClO鈧)鈧 and H鈧侽.
06

Balance the molecular equation

The balanced molecular equation is: Cu(OH)鈧(s) + 2HClO鈧(aq) 鈫 Cu(ClO鈧)鈧(aq) + 2H鈧侽(l)
07

Write the complete ionic equation

Breaking each compound into its individual ions, we get: Cu(OH)鈧(s) + 2H鈦(aq) + 2ClO鈧勨伝(aq) 鈫 Cu虏鈦(aq) + 2ClO鈧勨伝(aq) + 2H鈧侽(l)
08

Write the net ionic equation

Cancel the spectator ions (2ClO鈧勨伝) and obtain the net ionic equation: Cu(OH)鈧(s) + 2H鈦(aq) 鈫 Cu虏鈦(aq) + 2H鈧侽(l) (c) Al(OH)鈧(s) + HNO鈧(aq) 鈫
09

Predict the products

We have an acid-base reaction with Al鲁鈦 and 3OH鈦 reacting with H鈦 and NO鈧冣伝 to form Al(NO鈧)鈧 and H鈧侽.
10

Balance the molecular equation

The balanced molecular equation is: Al(OH)鈧(s) + 3HNO鈧(aq) 鈫 Al(NO鈧)鈧(aq) + 3H鈧侽(l)
11

Write the complete ionic equation

Breaking each compound into its individual ions, we get: Al(OH)鈧(s) + 3H鈦(aq) + 3NO鈧冣伝(aq) 鈫 Al鲁鈦(aq) + 3NO鈧冣伝(aq) + 3H鈧侽(l)
12

Write the net ionic equation

Cancel the spectator ions (3NO鈧冣伝) and obtain the net ionic equation: Al(OH)鈧(s) + 3H鈦(aq) 鈫 Al鲁鈦(aq) + 3H鈧侽(l)

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Most popular questions from this chapter

You know that an unlabeled bottle contains a solution of one of the following: \(\mathrm{AgNO}_{3}, \mathrm{CaCl}_{2}\), or \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) A friend suggests that you test a portion of the solution with \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and then with \(\mathrm{NaCl}\) solutions. Explain how these two tests together would be sufficient to determine which salt is present in the solution.

(a) Calculate the molarity of a solution made by dissolving \(0.750\) grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in enough water to form exactly \(850 \mathrm{~mL}\) of solution. (b) How many moles of \(\mathrm{KMnO}_{4}\) are present in \(250 \mathrm{~mL}\) of a \(0.0475 \mathrm{M}\) solution? (c) How many milliliters of \(11.6 \mathrm{M} \mathrm{HCl}\) solution are needed to obtain \(0.250 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

(a) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of some salt and then spill some of it. What happens to the concentration of the solution left in the container? (b) Suppose you prepare \(500 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) aqueous solution of some salt and let it sit out, uncovered, for a long time, and some water evaporates. What happens to the concentration of the solution left in the container? (c) \(A\) certain volume of a \(0.50 \mathrm{M}\) solution contains \(4.5 \mathrm{~g}\) of a salt. What mass of the salt is present in the same volume of a \(2.50 \mathrm{M}\) solution?

A sample of solid \(\mathrm{Ca}(\mathrm{OH})_{2}\) is stirred in water at \(30^{\circ} \mathrm{C}\) until the solution contains as much dissolved \(\mathrm{Ca}(\mathrm{OH})_{2}\) as it can hold. A \(100-\mathrm{mL}\) sample of this solution is withdrawn and titrated with \(5.00 \times 10^{-2} \mathrm{M} \mathrm{HBr}\). It requires \(48.8 \mathrm{~mL}\) of the acid solution for neutralization. What is the molarity of the \(\mathrm{Ca}(\mathrm{OH})_{2}\) solution? What is the solubility of \(\mathrm{Ca}(\mathrm{OH})_{2}\) in water, at \(30^{\circ} \mathrm{C}\), in grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) per \(100 \mathrm{~mL}\) of solution?

You are presented with three white solids, \(A, B\), and \(C\), which are glucose (a sugar substance), \(\mathrm{NaOH}\), and AgBr. Solid A dissolves in water to form a conducting solution. B is not soluble in water. C dissolves in water to form a nonconducting solution. Identify \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). [Section 4.2]
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