Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 129

A 450.0 -mL sample of a \(0.257-M\) solution of silver nitrate is mixed with 400.0 \(\mathrm{mL}\) of 0.200\(M\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?

Short Answer

Expert verified
The concentration of \(\text{Cl}^-\) in the solution after the reaction is complete is \(0.18824 \frac{\text{mol}}{\text{L}}\).

Step by step solution

01

Calculate moles of each reactant

First, we need to calculate the moles of each reactant. To do this, we multiply the molarity of each solution by its volume in liters: Moles of \(\text{AgNO}_3\) \(= 0.257 \frac{\text{mol}}{\text{L}} \cdot 0.450 \text{L} = 0.11565 \text{mol}\) Moles of \(\text{CaCl}_2\) \(= 0.200 \frac{\text{mol}}{\text{L}} \cdot 0.400 \text{L} = 0.08000 \text{mol}\)
02

Determine the limiting reactant

Next, we need to determine the limiting reactant. According to the stoichiometry of the balanced equation, we need 2 moles of \(\text{AgNO}_3\) to react with 1 mole of \(\text{CaCl}_2\). So, we compare the ratio of their moles: Ratio \( = \frac{0.11565}{2 \cdot 0.08000}= 0.72313 \) Since the ratio is greater than 1, we have an excess of \(\text{AgNO}_3\). Therefore, \(\text{CaCl}_2\) is the limiting reactant.
03

Calculate the amount of \(\text{Cl}^-\) after the reaction

Based on stoichiometry, 1 mole of \(\text{CaCl}_2\) produces 2 moles of \(\text{Cl}^-\), so after the reaction is complete, we will have: Moles of \(\text{Cl}^-\) \(= 2 \cdot 0.08000 \text{mol} = 0.16000 \text{mol}\)
04

Calculate the total volume of the final solution

The total volume of the final solution is the sum of the volumes of both solutions: Total volume \(= 0.450 \text{L} + 0.400 \text{L} = 0.850 \text{L}\)
05

Calculate the final concentration of \(\text{Cl}^-\)

To find the final concentration of \(\text{Cl}^-\), we divide the moles of \(\text{Cl}^-\) by the total volume of the final solution: Final concentration of \(\text{Cl}^-\) \(= \frac{0.16000 \text{mol}}{0.850 \text{L}} = 0.18824 \frac{\text{mol}}{\text{L}}\) Therefore, the concentration of \(\text{Cl}^-\) in the solution after the reaction is complete is \(0.18824 \frac{\text{mol}}{\text{L}}\).

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 solution of permanganate is standardized by titration with oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right) .\) It required 28.97 \(\mathrm{mL}\) of the permanganate solution to react completely with 0.1058 g of oxalic acid. The unbalanced equation for the reaction is $$\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \stackrel{\mathrm{Acidic}}{\longrightarrow} \mathrm{Mn}^{2+}(a q)+\mathrm{CO}_{2}(g)$$ What is the molarity of the permanganate solution?

What volume of 0.0521\(M \mathrm{Ba}(\mathrm{OH})_{2}\) is required to neutralize exactly 14.20 \(\mathrm{mL}\) of 0.141 $\mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4} ?$ Phosphoric acid contains three acidic hydrogens.

The iron content of iron ore can be determined by titration with a standard \(\mathrm{KMnO}_{4}\) solution. The iron ore is dissolved in \(\mathrm{HCl},\) and all the iron is reduced to \(\mathrm{Fe}^{2+}\) ions. This solution is then titrated with \(\mathrm{KMnO}_{4}\), solution, producing \(\mathrm{Fe}^{3+}\) and \(\mathrm{Mn}^{2+}\) ions in acidic solution. If it required 38.37 \(\mathrm{mL}\) of 0.0198 \(\mathrm{M}\) \(\mathrm{KMnO}_{4}\) to titrate a solution made from 0.6128 \(\mathrm{g}\) of iron ore, what is the mass percent of iron in the iron ore?

Tris(pentafluorophenyl)borane, commonly known by its acronym BARF, is frequently used to initiate polymerization of ethylene or propylene in the presence of a catalytic transition metal compound. It is composed solely of C, F, and B; it is 42.23% C and 55.66% F by mass. a. What is the empirical formula of BARF? b. A 2.251-g sample of BARF dissolved in 347.0 mL of solution produces a 0.01267-M solution. What is the molecular formula of BARF?

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

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Organic 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