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Chapter 4: Problem 67

Write the balanced formula equation for the acid-base reactions that occur when the following are mixed. a. potassium hydroxide (aqueous) and nitric acid b. barium hydroxide (aqueous) and hydrochloric acid c. perchloric acid \(\left[\mathrm{HClO}_{4}(a q)\right]\) and solid iron(III) hydroxide d. solid silver hydroxide and hydrobromic acid e. aqueous strontium hydroxide and hydroiodic acid

Short Answer

Expert verified
a. KOH + \(\mathrm{HNO}_{3}\) 鈫� H2O + KNO3 b. \(\mathrm{Ba(OH)}_{2}\) + 2HCl 鈫� 2H2O + BaCl2 c. 3HClO4 + \(\mathrm{Fe(OH)}_{3}\) 鈫� 3H2O + Fe(ClO4)3 d. \(\mathrm{AgOH}\) + HBr 鈫� H2O + AgBr e. \(\mathrm{Sr(OH)}_{2}\) + 2HI 鈫� 2H2O + SrI2

Step by step solution

01

a. potassium hydroxide (aqueous) and nitric acid

1. Write the formulas of the reactants: potassium hydroxide (KOH) and nitric acid (\(\mathrm{HNO}_{3}\)). 2. Identify the products of the reaction: When an acid (H+) reacts with a base (OH-), they form water (H2O) and a salt. In this case, the salt formed is potassium nitrate (KNO3). 3. Write the unbalanced equation: KOH + \(\mathrm{HNO}_{3}\) 鈫� H2O + KNO3 4. Balance the equation: The equation is already balanced. The balanced formula equation for the reaction is: KOH + \(\mathrm{HNO}_{3}\) 鈫� H2O + KNO3
02

b. barium hydroxide (aqueous) and hydrochloric acid

1. Write the formulas of the reactants: barium hydroxide (\(\mathrm{Ba(OH)}_{2}\)) and hydrochloric acid (\(\mathrm{HCl}\)). 2. Identify the products of the reaction: Water (H2O) and the salt barium chloride (BaCl2). 3. Write the unbalanced equation: \(\mathrm{Ba(OH)}_{2}\) + HCl 鈫� H2O + BaCl2 4. Balance the equation: To balance the equation, we need 2 moles of hydrochloric acid to provide enough H+ ions for both OH- ions from barium hydroxide: \(\mathrm{Ba(OH)}_{2}\) + 2HCl 鈫� 2H2O + BaCl2 The balanced formula equation for the reaction is: \(\mathrm{Ba(OH)}_{2}\) + 2HCl 鈫� 2H2O + BaCl2
03

c. perchloric acid and solid iron(III) hydroxide

1. Write the formulas of the reactants: Perchloric acid (HClO4) and solid iron(III) hydroxide (\(\mathrm{Fe(OH)}_{3}\)). 2. Identify the products of the reaction: Water (H2O) and the salt iron(III) perchlorate (Fe(ClO4)3). 3. Write the unbalanced equation: HClO4 + \(\mathrm{Fe(OH)}_{3}\) 鈫� H2O + Fe(ClO4)3 4. Balance the equation: To balance the equation, we need 3 moles of perchloric acid: 3HClO4 + \(\mathrm{Fe(OH)}_{3}\) 鈫� 3H2O + Fe(ClO4)3 The balanced formula equation for the reaction is: 3HClO4 + \(\mathrm{Fe(OH)}_{3}\) 鈫� 3H2O + Fe(ClO4)3
04

d. solid silver hydroxide and hydrobromic acid

1. Write the formulas of the reactants: Solid silver hydroxide (\(\mathrm{AgOH}\)) and hydrobromic acid (\(\mathrm{HBr}\)). 2. Identify the products of the reaction: Water (H2O) and the salt silver bromide (AgBr). 3. Write the unbalanced equation: \(\mathrm{AgOH}\) + HBr 鈫� H2O + AgBr 4. Balance the equation: The equation is already balanced. The balanced formula equation for the reaction is: \(\mathrm{AgOH}\) + HBr 鈫� H2O + AgBr
05

e. aqueous strontium hydroxide and hydroiodic acid

1. Write the formulas of the reactants: Aqueous strontium hydroxide (\(\mathrm{Sr(OH)}_{2}\)) and hydroiodic acid (\(\mathrm{HI}\)). 2. Identify the products of the reaction: Water (H2O) and the salt strontium iodide (SrI2). 3. Write the unbalanced equation: \(\mathrm{Sr(OH)}_{2}\) + HI 鈫� H2O + SrI2 4. Balance the equation: To balance the equation, we need 2 moles of hydroiodic acid: \(\mathrm{Sr(OH)}_{2}\) + 2HI 鈫� 2H2O + SrI2 The balanced formula equation for the reaction is: \(\mathrm{Sr(OH)}_{2}\) + 2HI 鈫� 2H2O + SrI2

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is essential for accurately representing a chemical reaction. Each side of the equation must have equal numbers of each type of atom. This balance reflects the conservation of mass, meaning matter isn't created or destroyed.
  • Start by writing the unbalanced equation with all reactants and products.
  • Count the atoms of each element on both sides of the equation.
  • Adjust coefficients (the numbers in front of compounds) to balance the atoms.
For example, when barium hydroxide reacts with hydrochloric acid, the equation becomes \( \mathrm{Ba(OH)}_{2} + 2\mathrm{HCl} \rightarrow 2\mathrm{H}_{2}\mathrm{O} + \mathrm{BaCl}_{2} \). Here, we added a coefficient of 2 in front of \( \mathrm{HCl} \) to balance the hydrogen and chlorine atoms. Each oxygen and barium atom is already balanced, making the entire equation consistent with the laws of chemistry.
Neutralization Reactions
Neutralization reactions occur when an acid and a base react to form water and a salt, neutralizing each other's effects. These are an important part of chemistry, often used in everyday life.
  • An acid contributes \( \mathrm{H}^{+} \) ions, while a base provides \( \mathrm{OH}^{-} \) ions.
  • When they combine, they form water \( (\mathrm{H}_2\mathrm{O}) \).
  • The remaining ions form a salt, which is the other product of the reaction.
For instance, when nitric acid is neutralized by potassium hydroxide, we get potassium nitrate and water. The equation \( \mathrm{KOH} + \mathrm{HNO}_{3} \rightarrow \mathrm{H}_{2}\mathrm{O} + \mathrm{KNO}_{3} \) showcases this process. Neutralization reactions help in managing pH levels in solutions, which is crucial for various chemical applications.
Products of Acid-Base Reactions
In acid-base reactions, the products are typically a salt and water. Understanding these products helps predict the outcomes of such reactions in practical situations.
  • The salt is formed from the cation of the base and the anion of the acid.
  • Water is produced from the \( \mathrm{H}^{+} \) ion of the acid and the \( \mathrm{OH}^{-} \) ion of the base.
For example, when solid silver hydroxide reacts with hydrobromic acid, silver bromide and water are formed: \( \mathrm{AgOH} + \mathrm{HBr} \rightarrow \mathrm{H}_{2}\mathrm{O} + \mathrm{AgBr} \). The silver from the hydroxide and bromine from the acid combine to form the salt, silver bromide, while water is produced as a neutralizing agent. These predictable products play a key role in fields like pharmaceuticals, agriculture, and environmental science.

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Most popular questions from this chapter

Chlorisondamine chloride \(\left(\mathrm{C}_{4} \mathrm{H}_{20} \mathrm{Cl}_{6} \mathrm{~N}_{2}\right)\) is a drug used in the treatment of hypertension. A \(1.28-\mathrm{g}\) sample of a medication containing the drug was treated to destroy the organic material and to release all the chlorine as chloride ion. When the filtered solution containing chloride ion was treated with an excess of silver nitrate, \(0.104 \mathrm{~g}\) silver chloride was recovered. Calculate the mass percent of chlorisondamine chloride in the medication, assuming the drug is the only source of chloride.

Saccharin \(\left(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\right)\) is sometimes dispensed in tablet form. Ten tablets with a total mass of \(0.5894 \mathrm{~g}\) were dissolved in water. The saccharin was oxidized to convert all the sulfur to sulfate ion, which was precipitated by adding an excess of barium chloride solution. The mass of \(\mathrm{BaSO}_{4}\) obtained was \(0.5032 \mathrm{~g}\). What is the average mass of saccharin per tablet? What is the average mass percent of saccharin in the tablets?

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The ion formed as a product, \(\mathrm{X}^{2-}\), was shown to have 36 total electrons. What is element \(\mathrm{X}\) ? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\). To completely neutralize a sample of \(\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{~mL}\) of \(0.175 \mathrm{M}\) \(\mathrm{OH}^{-}\) solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

What mass of barium sulfate can be produced when \(100.0 \mathrm{~mL}\) of a \(0.100-M\) solution of barium chloride is mixed with \(100.0 \mathrm{~mL}\) of a \(0.100-M\) solution of iron(III) sulfate?

What volume of each of the following bases will react completely with \(25.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) ? a. \(0.100 \mathrm{M} \mathrm{NaOH}\) b. \(0.0500 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\) c. \(0.250 \mathrm{M} \mathrm{KOH}\)
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