HomeworkLib
Question

1. A saturated solution of lead(II) chloride was prepared by dissolving PbCl2 solid in water. The concentration of Pb+2...

1. A saturated solution of lead(II) chloride was prepared by dissolving PbCl2 solid in water. The concentration of Pb+2 ion in the solution was found to be 1.62*10^-2M . Calculate Ksp for PbCl2 .

2. The value of Ksp for silver chromate, Ag2CrO4 is 9.0*10^-12 . Calculate the solubility of Ag2CrO4 in grams per liter.
science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1
Concepts and reason

The concepts used here are based on the solubility product constant and molar solubility.

First, the solubility product of lead chloride is calculated. After that, using the solubility product constant of substance the molar solubility or concentration of silver chromate is calculated.

Fundamentals

Solubility product:

The solubility product helps to determine the solubility of the compound in water.

Higher the value of solubility product, higher will be the solubility of that compound in water. It tells the level of solute dissolves in a particular solution.

Example-

AB, aA +bB

The expression for the solubility product will be shown below.

K. =[A] [B] …… (1)

Here, is the concentration of andis the Concentration of , a and b are the coefficients of the respective ion.

Molar solubility:

The solubility (molar solubility) of a solid is expressed as the concentration of the dissolved solid in a saturated solution.

The molar solubility can be calculated as shown below.

aA (s) bB(aq)+cC(aq) 0 0 K =xxx

x= K …… (2)

Here, is the molar solubility.

(1)

The balanced reaction of dissolution of lead chloride is as follows.

PÉCI, (s) Pb²+ (aq) + 2CF (aq)

The ICE table of dissociation of leadchloride is shown below.

PbCl2 (s) Pb2+ (aq) + 2C1 (aq) 0 Intial concentration change in concentration Equilbrium concentration -x +2x 2x

The solubility product expression can be written as follows.

<(+)(x)= dy {-17[1zqa] = y

To calculate the of РЬСІ, substitute the value of as 1.62x10 PM in the above equation.

K. =(1.62x10-²M)(2x1.62x10-?M) K.=1.62x10-2x(3.24x10-2) Ko = 1.62x10-2x10.49x10-4 K =1.7x10-5

(2)

The reaction of silver chromate in water is shown below.

Ag, Cro (s) 2Ag* (aq) + Cro - (aq)

The ICE table of dissociation of silver chromate is shown below.

Ag,Cro, (s) 2Ag* (aq) + Cro,?- (aq) 0 Intial concentration change in concentration Equilbrium concentration +2x 2x

The expression for solubility product for silver chromate will be shown below.

[+*o-ɔ][.8v]= *y

Substitute 9.0x10-12 as the value of and as the value of [agt and as the value of [Cro ²-1 .

9.0x10-12 = (2x)” (x) 9.0x10-12 4 2.25x10-12 = x 1.310x104 mol L=x

Conversion of the unit of solubility from mol per liter to gram per liter is as shown below.

x (g.L-)=Mxx (mol.L-) …… (3)

Substitute 1.310x104 mol as the value of x(mol.L) and 331.73 g mol as the value of the molar mass of silver nitrate (M) in equation (3)

x = 331.73 g.mot x1.310x104 mol - x = 4.34x102 g

Ans: Part 1

The solubility product of lead chloride (PbCI,) is 1.7x10-5 .

2 0
Add a comment
Know the answer?
Add Answer to:
1. A saturated solution of lead(II) chloride was prepared by dissolving PbCl2 solid in water. The concentration of Pb+2...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • Chapter 15 Question 9 1)A saturated solution of lead(II) chloride, PbCl2, was prepared by dissolving solid PbCl2 in wa...

    Chapter 15 Question 9 1)A saturated solution of lead(II) chloride, PbCl2, was prepared by dissolving solid PbCl2 in water. The concentration of Pb2+ ion in the solution was found to be 1.62×10−2 M . Calculate Ksp for PbCl2. 2)The value of Ksp for silver sulfate, Ag2SO4, is 1.20×10−5. Calculate the solubility of Ag2SO4 in grams per liter.

  • A saturated solution of barium fluoride, BaF2, was prepared by dissolving solid BaF2 in water. The...

    A saturated solution of barium fluoride, BaF2, was prepared by dissolving solid BaF2 in water. The concentration of Ba2+ ion in the solution was found to be 7.52×10−3 M . Calculate Ksp for BaF2. The value of Ksp for silver chromate, Ag2CrO4, is 9.0×10−12. Calculate the solubility of Ag2CrO4 in grams per liter.  

  • Part A A saturated solution of barium fluoride, BaF2, was prepared by dissolving solid BaF2 in water. The concentration...

    Part A A saturated solution of barium fluoride, BaF2, was prepared by dissolving solid BaF2 in water. The concentration of Ba2+ ion in the solution was found to be 7.52�10?3M . Calculate Ksp for BaF2. Part B The value of Ksp for silver carbonate, Ag2CO3, is 8.10�10?12. Calculate the solubility of Ag2CO3 in grams per liter.

  • A Calculate the molar solubility of PbCl2 in a 0.2340 M lead(II) perchlorate, Pb(ClO )2 solution...

    A Calculate the molar solubility of PbCl2 in a 0.2340 M lead(II) perchlorate, Pb(ClO )2 solution Solubility = B. Let's say we have a beaker where a saturated solution of lead(II) chloride is in equilibrium with solid lead(II) chloride in which of these cases will the molar solubility be lowest after equilibrium is reestablished? After the addition of solid NaNO3. After the addition of 0.120 moles of Clion. Not enough information given, After letting some of the solvent evaporate None...

  • #5 Write the solubility product expression for PbCl2. Using the concentration for the Pb+2 and Cl-...

    #5 Write the solubility product expression for PbCl2. Using the concentration for the Pb+2 and Cl- ions, solve for your experimental Ksp. #6 Using your book, find the theoretical Ksp for PbCl2 to determine your percent error A Solubility Product Constant Introduction: Many substances are very soluble in water. However, in this experiment you will be concerned with substances that are insoluble or only slightly soluble. Dynamic equilibrium is established when an excess of a slightly soluble substance is placed...

  • Please help asap =) 2. Lead(II)chloride is insoluble in water Kap PbCn 1.7x10-) and silver chloride...

    Please help asap =) 2. Lead(II)chloride is insoluble in water Kap PbCn 1.7x10-) and silver chloride is very insoluble in water (Ksp Agci 1.8x10-10). The reactions when they go into solution are: AgCl(s) ←→ Ag+(aq) + Cl-(aq) PbCl2(s)艹Pb+2(aq) + 2 Cl-(aq) a. What are the expressions for the equilibrium constants, Kop for the above reactions? b. Explain why it is possible to dissolve more lead(I)chloride in solutions in which the concentration of the silver ion is present. c. Explain what...

  • Silver chloride, AgCl, is a sparingly soluble solid. Answer the following questions about a saturated solution...

    Silver chloride, AgCl, is a sparingly soluble solid. Answer the following questions about a saturated solution prepared by placing solid silver chloride in a 1.94 x 10-5 M NaCl(aq) solution. At some temperature, the silver ion concentration, [Ag+], was found to be 6.24 x 10-6 M. (a) What is the concentration of chloride ions, [CI – ], in the resulting solution? XM (b) What is the molar solubility of silver chloride, AgCl, in 1.94 x 10-5 M NaCl? 4.9 6.24e-6...

  • A saturated solution of magnesium fluoride , MgF2, was prepared by dissolving solid MgF2 in water. The concentration of...

    A saturated solution of magnesium fluoride , MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 1.18

  • Learning Goal: To learn how to calculate the solubility from Kspand vice versa. Consider the following equilibrium be...

    Learning Goal: To learn how to calculate the solubility from Kspand vice versa. Consider the following equilibrium between a solid salt and its dissolved form (ions) in a saturated solution: CaF2(s)⇌Ca2+(aq)+2F−(aq) At equilibrium, the ion concentrations remain constant because the rate of dissolution of solid CaF2 equals the rate of the ion crystallization. The equilibrium constant for the dissolution reaction is Ksp=[Ca2+][F−]2 Ksp is called the solubility product and can be determined experimentally by measuring thesolubility, which is the amount...

  • 1.. A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate...

    1.. A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 29.0 mL of NaOH. 2. A saturated solution of magnesium fluoride, MgF2, was prepared by dissolving solid MgF2 in water. The concentration of Mg2+ ion in the solution was found to be 1.18×10−3 M . Calculate Ksp for MgF2. 3. The value of Ksp for silver carbonate, Ag2CO3, is 8.10×10−12. Calculate the solubility of Ag2CO3 in grams per...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT