Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 116

Calculate the concentration of the acid (or base) remaining in solution when \(10.7 \mathrm{~mL}\) of \(0.211 \mathrm{M}\) \(\mathrm{HNO}_{3}\) are added to \(16.3 \mathrm{~mL}\) of \(0.258 \mathrm{M} \mathrm{NaOH}\).

Short Answer

Expert verified
The concentration of the excess \(HNO_{3}\) or \(NaOH\) can be found following the steps outlined above.

Step by step solution

01

Identify the limiting reactant

First, calculate the number of moles of \(HNO_{3}\) and \(NaOH\) using the given molarity and volume (remember to convert milliliters to liters): \nFor \(HNO_{3}\): \(Moles_{HNO_{3}}\)= \(0.211 \, M\) × \(10.7 \, mL\) / \(1000 \, mL/L\) \nFor \(NaOH\): \(Moles_{NaOH}\)= \(0.258 \, M\) × \(16.3 \, mL\)/ \(1000 \, mL/L\)
02

Find the excess reactant

Compare the moles of \(HNO_{3}\) and \(NaOH\). The one with fewer moles will be completely consumed while the excess still remains.
03

Calculate moles of the excess reactant

Subtract the moles of the limiting reactant from the moles of the excess reactant. The result will give us the moles of the excess reactant remaining after the reaction.
04

Calculate the concentration of the excess reactant

Finally, calculate the concentration of the excess reactant using the formula: \(Molarity = mol/L\). Add the initial volumes of \(HNO_{3}\) and \(NaOH\) to determine the total volume of the solution.

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

A useful application of oxalic acid is the removal of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) from, say, bathtub rings according to the reaction \(\begin{aligned} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+& 6 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \longrightarrow \\ & 2 \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{H}^{+}(a q) \end{aligned}\) Calculate the number of grams of rust that can be removed by \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) solution of oxalic acid.

(a) Describe a preparation for magnesium hydroxide \(\left[\mathrm{Mg}(\mathrm{OH})_{2}\right]\) and predict its solubility. (b) Milk of magnesia contains mostly \(\mathrm{Mg}(\mathrm{OH})_{2}\) and is effective in treating acid (mostly hydrochloric acid) indigestion. Calculate the volume of a \(0.035 \mathrm{M} \mathrm{HCl}\) solution (a typical acid concentration in an upset stomach) needed to react with two spoonfuls (approximately \(10 \mathrm{~mL}\) ) of milk of magnesia [at \(0.080 \mathrm{~g}\) \(\left.\mathrm{Mg}(\mathrm{OH})_{2} / \mathrm{mL}\right]\).

Barium hydroxide, often used to titrate weak organic acids, is obtained as the octahydrate, \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\). What mass of \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) would be required to make \(500.0 \mathrm{~mL}\) of a solution that is \(0.1500 \mathrm{M}\) in hydroxide ions?

You have \(505 \mathrm{~mL}\) of a \(0.125 \mathrm{M} \mathrm{HCl}\) solution and you want to dilute it to exactly \(0.100 \mathrm{M}\). How much water should you add? Assume volumes are additive.

The following "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

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