{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

RDGTitrationStoichtxt

# RDGTitrationStoichtxt - brady_c04_127-174v2.1 1:18 PM Page...

This preview shows pages 1–2. Sign up to view the full content.

4.8 Chemical Analysis and Titration Are Applications of Solution Stoichiometry 163 A solution of AgNO 3 is added to a solution of CaCl 2 , producing a precipitate of AgCl. (Andy Washnik.) Simple reasoning works well here. Since the coefficients of Ag 1 and Cl 2 are the same, the numbers of moles that react must be equal. To solve the problem, the first step will be to calculate the molarities of the ions in the so- lutions being mixed. Next, using the volume and molarity of the Cl 2 solution, we will calculate the moles of Cl 2 available. Then we’ll use the coefficients of the equation to find the moles of Ag 1 that react. Finally, we’ll use the molarity of the Ag 1 solution to determine the volume of the 0.100 M AgNO 3 solution needed. SOLUTION: We begin by finding the concentrations of the ions in the reacting solutions: 0.100 M AgNO 3 contains 0.100 M Ag 1 and 0.100 M NO 3 2 0.400 M CaCl 2 contains 0.400 M Ca 2 1 and 0.800 M Cl 2 We’re only interested in the Ag 1 and Cl 2 ; the Ca 2 1 and NO 3 2 are spectator ions. For our pur- poses, then, the solution concentrations are 0.100 M Ag 1 and 0.800 M Cl 2 . Having these val- ues, we can now restate the problem: How many milliliters of 0.100 M Ag 1 solution are needed to react completely with 25.0 mL of 0.800 M Cl 2 solution? The moles of Cl 2 available for reaction are obtained from the molarity and volume (0.0250 L) of the Cl 2 solution. The coefficients of the equation tell us the Ag 1 and Cl 2 combine in a 1-to-1 mole ratio, so . Finally, we calculate the volume of the Ag 1 solution us- ing its molarity as a conversion factor. As we’ve done earlier, we use the factor that makes the units cancel correctly. Our calculations tell us that we must use 0.200 L or 200 mL of the AgNO 3 solution. We could also have used the following chain calculation, of course. 0.0200 mol Ag + * 1.00 L Ag + soln 0.100 mol Ag + = 0.200 L Ag + soln 0.0200 mol Cl - 3 0.0200 mol Ag + 0.0250 L Cl - soln * 0.800 mol Cl - 1.00 L Cl - soln = 0.0200 mol Cl - 25.0 mL Cl - soln 3 ? mL Ag + soln 0.0250 L Cl - soln * 0.800 mol Cl - 1.00 L Cl - soln * 1 mol Ag + 1 mol Cl - * 1.00 L Ag + soln 0.100 mol Ag + = 0.200 L Ag + soln IS THE ANSWER REASONABLE? The silver ion concentration is one-eighth as large as the chloride ion concentration. Since the ions react one-for-one, we will need eight times as much silver ion solution as chloride solution. Eight times the amount of chloride solution, 25 mL, is 200 mL, which is the answer we obtained. Therefore, the answer appears to be correct. Practice Exercise 35: Suppose 18.4 mL of 0.100 M AgNO 3 solution was needed to react completely with 20.5 mL of CaCl 2 solution. What is the molarity of the CaCl 2 solu- tion? Use the net ionic equation in the preceding example to work the problem. (Hint: How can you calculate molarity from moles and volume, and how can you calculate the molarity of the CaCl 2 solution from the molarity of Cl 2 ?) Practice Exercise 36: How many milliliters of 0.500 M KOH are needed to react completely with 60.0 mL of 0.250 M FeCl 2 solution to precipitate Fe(OH) 2 ? The net ionic equation is 4.8 C HEMICAL ANALYSIS AND TITRATION ARE APPLICATIONS OF SOLUTION STOICHIOMETRY Chemical analyses fall into two categories. In a qualitative analysis

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

RDGTitrationStoichtxt - brady_c04_127-174v2.1 1:18 PM Page...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online