Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 115

A 22.02 -mL solution containing \(1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with a 28.64 -mL solution containing \(1.073 \mathrm{~g}\) \(\mathrm{NaOH} .\) Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.

Short Answer

Expert verified
The concentration of \(Na^+\) and \(OH^-\) ions each is 0.0997M after the reaction.

Step by step solution

01

Calculate Initial Moles of Each Reactant

First, calculate the initial moles of each reactant, \(Mg(NO_3)_2\) and \(NaOH\), using their masses and molar masses. The molar mass of \(Mg(NO_3)_2\) is approximately 148.32 g/mol and of \(NaOH\) is approximately 40 g/mol. So, the moles of \(Mg(NO_3)_2\) and \(NaOH\) can be calculated as follows: Moles of \(Mg(NO_3)_2\) = \(1.615g / 148.32g/mol = 0.01089 mol\)Moles of \(NaOH\) = \(1.073g / 40g/mol = 0.02683 mol\)
02

Identify the Limiting Reagent

The balanced chemical equation for the reaction between \(Mg(NO_3)_2\) and \(NaOH\) is: \(Mg(NO_3)_2 + 2NaOH \rightarrow Mg(OH)_2 + 2NaNO_3\). From this equation, we see that 1 mole of \(Mg(NO_3)_2\) reacts with 2 moles of \(NaOH\). From the actual moles of these reactants, \(Mg(NO_3)_2\) can react with \(2* 0.01089 = 0.02178 mol\) of \(NaOH\), while we originally have 0.02683 mol of \(NaOH\). Hence, \(Mg(NO_3)_2\) is the limiting reagent.
03

Calculate Moles of Ions Remaining

After the reaction, all the \(Mg(NO_3)_2\) reacts so the \(Mg^{2+}\) ion is not present. Meanwhile, \(NaOH\) still has remaining moles, which equals to the original moles minus the amount that reacted with \(Mg(NO_3)_2\). Thus, remaining moles of \(NaOH\) = \(0.02683mol - 0.02178mol = 0.00505 mol\). Since in the reaction, one mole of \(NaOH\) provides one mole of \(Na^+\) and \(OH^-\), we have remaining 0.00505 mol \(Na^+\) and \(OH^-\).
04

Calculate the Concentrations of the Remaining Ions

The total volume of the solution is \(22.02mL + 28.64mL = 50.66mL = 0.05066L\). The concentration can be calculated by dividing the moles of ions with the volume of the solution in liters. Thus, the concentrations of the remaining ions are:Concentration of \(Na^+\) = \(0.00505mol / 0.05066L = 0.0997M\)Concentration of \(OH^-\) = \(0.00505mol / 0.05066L = 0.0997M\)

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

Describe the basic steps involved in gravimetric analysis. How does this procedure help us determine the identity of a compound or the purity of a compound if its formula is known?

On the basis of oxidation number considerations, one of the following oxides would not react with molecular oxygen: \(\mathrm{NO}, \mathrm{N}_{2} \mathrm{O}, \mathrm{SO}_{2}, \mathrm{SO}_{3}, \mathrm{P}_{4} \mathrm{O}_{6}\). Which one is it? Why?

How does an acid-base indicator work?

Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) is an important industrial chemical used in fertilizers, detergents, and in the food industry. It is produced by two different methods. In the electric furnace method elemental phosphorus \(\left(\mathrm{P}_{4}\right)\) is burned in air to form \(\mathrm{P}_{4} \mathrm{O}_{10}\), which is then reacted with water to give \(\mathrm{H}_{3} \mathrm{PO}_{4}\). In the wet process the mineral phosphate rock \(\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\right]\) is reacted with sulfuric acid to give \(\mathrm{H}_{3} \mathrm{PO}_{4}\) ( and HF and \(\mathrm{CaSO}_{4}\) ). Write equations for these processes and classify each step as precipitation, acid-base, or redox reaction.

How many grams of \(\mathrm{NaCl}\) are required to precipitate most of the \(\mathrm{Ag}^{+}\) ions from \(2.50 \times 10^{2} \mathrm{~mL}\) of \(0.0113 \mathrm{MAgNO}_{3}\) solution? Write the net ionic equation for the reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

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