c. When 9.5 g of an impure Silver Nitrate (AGNO, mol. wt 170 g/mol) was treated with excess sodium chloride (NaCl), 7.15 g of a precipitate (mol wt. 143 g/mol) was obtained. How many grams of AgNO3 was in the mixture? Only number needed Move To...

Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter3: Composition Of Substances And Solutions
Section: Chapter Questions
Problem 64E: What is the final concentration of the solution produced when 225.5 mL of a 0.09988-M solution of...
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c. When 9.5 g of an impure Silver Nitrate (AGNO3, mol. wt 170 g/mol) was treated with excess
sodium chloride (NaCI), 7.15 g of a precipitate (mol wt. 143 g/mol) was obtained. How many
grams of AgNO, was in the mixture? Only number needed
Move To...
Transcribed Image Text:c. When 9.5 g of an impure Silver Nitrate (AGNO3, mol. wt 170 g/mol) was treated with excess sodium chloride (NaCI), 7.15 g of a precipitate (mol wt. 143 g/mol) was obtained. How many grams of AgNO, was in the mixture? Only number needed Move To...
2. If it took 50 ml of 0.1Molar NaOH to reach the first equivalence point in the titration of 50 ml of
Carbonic acid whose Ka, is 10+ and Kaz is 10-".
(a) What is the concentration of the Acid?
D.l
1.
20
%3D
しR Lers
0.005
BasE pUN
(b) What is the pH when 25 ml of NaOH has been added?
PH=14-0.01=13.93
50+25= 15 → 0.0 15
Ool (0.75) = 1.5
0.005
M→ 0.1-0.07
105
POH
(c) What is the pH at the first equivalence point?
(d) What volume of NaOH is needed to reach the second equivalence point and what species exist in
solition at that pH?
(c) What will the pH of the solution when 75 ml of NaOH has been added to the solution?
Transcribed Image Text:2. If it took 50 ml of 0.1Molar NaOH to reach the first equivalence point in the titration of 50 ml of Carbonic acid whose Ka, is 10+ and Kaz is 10-". (a) What is the concentration of the Acid? D.l 1. 20 %3D しR Lers 0.005 BasE pUN (b) What is the pH when 25 ml of NaOH has been added? PH=14-0.01=13.93 50+25= 15 → 0.0 15 Ool (0.75) = 1.5 0.005 M→ 0.1-0.07 105 POH (c) What is the pH at the first equivalence point? (d) What volume of NaOH is needed to reach the second equivalence point and what species exist in solition at that pH? (c) What will the pH of the solution when 75 ml of NaOH has been added to the solution?
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