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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

A solution is prepared by mixing 75.0 mL of 0.020 M BaCl2 and 125 mL of 0.040 M K2SO4. What are the concentrations of barium and sulfate ions in this solution? Assume only SO 4 2 ions (no HSO 4 ) are present.

Interpretation Introduction

Interpretation:

The final concentration of barium Ba2+(aq) and sulfate SO42(aq) has to be calculated.

Concept introduction:

The formation of solid in a solution is known as precipitation.  Solubility product, Ksp is defined as the concentration of ions in a saturated solution where each ion is raised to the power of their coefficients. Ion product, Q is defined as the initial concentration of ions in any solution where each ion is raised to the power of their coefficients.

The relation between ion product, Q and solubility product, Ksp is as follows.

  • If Ksp>Q , then no precipitate will form.
  • If Ksp<Q , then precipitate will form.
  • If Ksp=Q , then the solution will be in just saturation.
Explanation

Explanation

To determine: The final concentration of Ba2+(aq) and SO42(aq) solution prepared by mixing 75mL of 0.020M BaCl2 to 125mL of 0.040M K2SO4 .

Given

Concentration of BaCl2 is 0.020M .

Concentration of K2SO4 is 0.040M .

Solubility product of BaC2O4 is 2.3×108 .

Volume of K2SO4 is 125mL .

Volume of BaCl2 is 75mL .

It is assumed that the concentration of Ba2+ and SO42(aq) is same as the concentration of BaCl2 and K2SO4 . Hence, concentration of Ba2+ and SO42(aq) is,

[Ba2+]=0.020M[SO42(aq)]=0.040M

The initial concentration of compound is calculated using the formula,

Initialconcentration=Volumeofcompound×Givenconcentration[Volumeoforiginalsolution(mL)+(Volumeofcompoundadded(mL))] (1)

Substitute the values of volume, given concentration of Ba2+ and volume of original solution in the above equation.

InitialconcentrationofBa2+=Volumeofcompound×Givenconcentration[Volumeoforiginalsolution(mL)+(Volumeofcompoundadded(mL))]=75mL(0.020)M75mL+125mL=7

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