Log out Go to App
Go to App

Learning Materials

Features

Discover

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. \(KOH (aq) + HNO3 (aq) \rightarrow H2O (l) + KNO3 (aq)\) b. \(Ba(OH)2 (aq) + 2 HCl (aq) \rightarrow 2 H2O (l) + BaCl2 (aq)\) c. \(3 HClO4 (aq) + Fe(OH)3 (s) \rightarrow 3 H2O (l) + Fe(ClO4)3 (aq)\) d. \(AgOH (s) + HBr (aq) \rightarrow H2O (l) + AgBr (s)\) e. \(Sr(OH)2 (aq) + 2 HI (aq) \rightarrow 2 H2O (l) + SrI2 (aq)\)

Step by step solution

01

a. Potassium hydroxide and nitric acid

For this reaction, the reactants are potassium hydroxide (KOH) and nitric acid (HNO3). In an acid-base reaction, the hydrogen (H+) from the acid will combine with the hydroxide (OH-) from the base to form water (H2O). The remaining ions will form a salt. In this case, the salt will be potassium nitrate (KNO3). So, we have: KOH (aq) + HNO3 (aq) → H2O (l) + KNO3 (aq) The equation is already balanced.
02

b. Barium hydroxide and hydrochloric acid

For this reaction, the reactants are barium hydroxide (Ba(OH)2) and hydrochloric acid (HCl). The acid-base reaction will produce water (H2O) and the salt barium chloride (BaCl2). The balanced equation is as follows: Ba(OH)2 (aq) + 2 HCl (aq) → 2 H2O (l) + BaCl2 (aq)
03

c. Perchloric acid and solid iron(III) hydroxide

This reaction involves perchloric acid (HClO4) and solid iron(III) hydroxide (Fe(OH)3). The products are water (H2O) and the salt iron(III) perchlorate (Fe(ClO4)3). The balanced equation is: 3 HClO4 (aq) + Fe(OH)3 (s) → 3 H2O (l) + Fe(ClO4)3 (aq)
04

d. Solid silver hydroxide and hydrobromic acid

The reactants in this case are solid silver hydroxide (AgOH) and hydrobromic acid (HBr). The products will be water (H2O) and the salt silver bromide (AgBr). The balanced equation is: AgOH (s) + HBr (aq) → H2O (l) + AgBr (s)
05

e. Aqueous strontium hydroxide and hydroiodic acid

Finally, we have the reaction between aqueous strontium hydroxide (Sr(OH)2) and hydroiodic acid (HI). The products will be water (H2O) and the salt strontium iodide (SrI2). The balanced equation is: Sr(OH)2 (aq) + 2 HI (aq) → 2 H2O (l) + SrI2 (aq)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from \(75.0 \mathrm{~mL}\) of a \(0.100 M\) solution of \(\mathrm{AgNO}_{3}\) ?

A \(1.42-\mathrm{g}\) sample of a pure compound, with formula \(\mathrm{M}_{2} \mathrm{SO}_{4}\), was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh \(1.36 \mathrm{~g}\). Determine the atomic mass of \(\mathrm{M}\), and identify \(\mathrm{M}\).

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-} \longrightarrow \mathrm{X}^{2-}+\mathrm{H}_{2} \mathrm{O} $$ 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?

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50 M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 M)\) sulfuric acid b. \(\mathrm{HCl}\) from "concentrated" \((12 \mathrm{M})\) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

Write the balanced formula, complete ionic, and net ionic equations for each of the following acid-base reactions. a. \(\mathrm{HClO}_{4}(a q)+\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightarrow\) b. \(\mathrm{HCN}(a q)+\mathrm{NaOH}(a q) \rightarrow\) c. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow\)
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free