Example / What volume of 0.40 M Ba(OH)2 must be added to 50 mL of 0.30 MNaOH to give a solution 0.50 M in OH.Solution:Let x= mL Ba (OH)2 and the final volume is (50+x) mLmmole OH- = mmole NaOH +2 x mmol Ba(OH)20.5 M x (50 + x)mL 0.30 M x 50 mL + 2 x (0.40 M x (x) mL)x = 33 mL Ba(0H)2Example / Prepare 500 mL of 0.010 M solution of Na from solid Na,CO3./-

1 mmol of Ba(OH)2 = 2 mmol of OH 1 mmol of NaOH = 1 mmol of OH Number of mmols = molarity x volume…

Example / What volume of 0.40 M Ba(OH)2 must be added to 50 mL of 0.30 M NAOH to give a solution 0.50 M in OH. Solution: Let x = mL Ba (OH)2 and the final volume is (50+x) mL mmole OH- = mmole Na0H + 2 x mmol Ba(OH)2 %3D 0.5M x (50 + x)mL = 0.30 M x 50 mL + 2 x (0.40MX (x) mL) x = 33 mL Ba(0H)2

Example / What volume of 0.40 M Ba(OH)2 must be added to 50 mL of 0.30 M NAOH to give a solution 0.50 M in OH. Solution: Let x = mL Ba (OH)2 and the final volume is (50+x) mL mmole OH- = mmole Na0H + 2 x mmol Ba(OH)2 %3D 0.5M x (50 + x)mL = 0.30 M x 50 mL + 2 x (0.40MX (x) mL) x = 33 mL Ba(0H)2

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Example / What volume of 0.40 M Ba(OH)2 must be added to 50 mL of 0.30 M
NaOH to give a solution 0.50 M in OH.
Solution:
Let x= mL Ba (OH)2 and the final volume is (50+x) mL
mmole OH- = mmole NaOH +2 x mmol Ba(OH)2
0.5 M x (50 + x)mL 0.30 M x 50 mL + 2 x (0.40 M x (x) mL)
x = 33 mL Ba(0H)2
Example / Prepare 500 mL of 0.010 M solution of Na from solid Na,CO3.
/-

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