(#17) A 189.0-mL aliquot of 0.390 M aqueous potassium hydroxide is mixed with 189.0 mL of 0.390 M aqueous magnesium nitrate.The precipitate is magnesium hydroxide.cWhat mass of precipitate is produced?Mass = ______________ gCalculate the concentration of each ion remaining in the solution after precipitation is complete. Assume that the precipitate is completely insoluble.Concentration of K+ = _________MConcentration of OH-= _________ MConcentration of MG2+ = _____________MConcentration of NO3- = ____________M

Answered: (#17) A 189.0-mL aliquot of 0.390 M…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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(#17) A 189.0-mL aliquot of 0.390 M aqueous potassium hydroxide is mixed with 189.0 mL of 0.390 M aqueous magnesium nitrate.

The precipitate is magnesium hydroxide.

c	What mass of precipitate is produced? Mass = ______________ g

Calculate the concentration of each ion remaining in the solution after precipitation is complete. Assume that the precipitate is completely insoluble.

Concentration of K+ = _________M

Concentration of OH-= _________ M

Concentration of MG2+ = _____________M

Concentration of NO3- = ____________M

Expert Solution

Step 1

A balanced chemical equation for the above reaction is represented as follows:

$2 K O H_{(a q)} + M g {(N O_{3})}_{2 (a q)} \to M g {(O H)}_{2 (s)} + 2 K N O_{3 (a q)}$

The precipitate is magnesium hydroxide.

$N u m b e r o f m o l e s o f K O H = 0.390 M \times 0.189 L = 0.07371 m o l$

$N u m b e r o f m o l e s o f M g {(N O_{3})}_{2} = 0.390 M \times 0.189 L = 0.07371 m o l$

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