constructing a BCA table, constructing an ICE table, writing the equilibrium constant expression, and use this information to determine the pH. The value of Ka for HC-H-O: is 1.5 x 10%. Complete Parts 1-4 before submitting your answer. NEXT > 40.0 mL of a 0.200 M HC-H-O: solution was titrated with 10.0 mL of 0.100 M Sr(OH): (a strong base). Fill in the BCA table with the appropriate value for each involved species to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of water in the reaction. HCH.O.(aq) + OH(aq) HO(1) . CHO (0) Before (mol) Change (mol) After (mol) 338 RESET 0.100 0.200 1.00 x 10- -1.00 × 10-³ 2.00 x 10- -2.00 x 10-³ 6.00 × 10 -6.00 × 10 7.00 × 10 -7.00 × 10°³ 8.00 × 10 -8.00 × 10 < PREV NEXT > Upon completion of the acid-base reaction, the C-H-O₂- ion is in equilibrium with water. Set up the ICE table in order to determine the unknown concentrations of reactants and products. HC.H-Ox(aq) + H:O(1) = HO-(aq) + CH-O:-(aq) Initial (M) Change (M) Equilibrium (M) RESET 0.100 0.120 0.140 0.0200 0.0400 0.100 + x 0.100 - x 0.120 + x 0.120-x 0.140+x 0.140-x 0.0200 + x 0.0200-x 0.0400+x 0.0400-x < PREV 2 3 4 NEXT > Based on your ICE table and the equilibrium expression for Ka, set up the expression for Ka in order to determine the unknown concentrations. Each reaction participant must be represented by one tile. Do not combine terms. Ka = = 1.5 × 10-5 RESET [0] [0.100] [0.120] [0.140] [0.0200] [0.0400] [x] [2x] [0.100 + x] [0.100-x] [0.120+x] [0.120-x] [0.140 + x] [0.140-x] [0.0200 + x] [0.0200 - x] [0.0400+x] [0.0400 - x] < PREV 2 3 Based on your ICE table and the equilibrium expression for Ka, determine the pH of this solution. pH = RESET 0 4.35 4.49 x 10-5 9.65 3.99 10.0 2.23 x 10-10 1.40 3.23
constructing a BCA table, constructing an ICE table, writing the equilibrium constant expression, and use this information to determine the pH. The value of Ka for HC-H-O: is 1.5 x 10%. Complete Parts 1-4 before submitting your answer. NEXT > 40.0 mL of a 0.200 M HC-H-O: solution was titrated with 10.0 mL of 0.100 M Sr(OH): (a strong base). Fill in the BCA table with the appropriate value for each involved species to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of water in the reaction. HCH.O.(aq) + OH(aq) HO(1) . CHO (0) Before (mol) Change (mol) After (mol) 338 RESET 0.100 0.200 1.00 x 10- -1.00 × 10-³ 2.00 x 10- -2.00 x 10-³ 6.00 × 10 -6.00 × 10 7.00 × 10 -7.00 × 10°³ 8.00 × 10 -8.00 × 10 < PREV NEXT > Upon completion of the acid-base reaction, the C-H-O₂- ion is in equilibrium with water. Set up the ICE table in order to determine the unknown concentrations of reactants and products. HC.H-Ox(aq) + H:O(1) = HO-(aq) + CH-O:-(aq) Initial (M) Change (M) Equilibrium (M) RESET 0.100 0.120 0.140 0.0200 0.0400 0.100 + x 0.100 - x 0.120 + x 0.120-x 0.140+x 0.140-x 0.0200 + x 0.0200-x 0.0400+x 0.0400-x < PREV 2 3 4 NEXT > Based on your ICE table and the equilibrium expression for Ka, set up the expression for Ka in order to determine the unknown concentrations. Each reaction participant must be represented by one tile. Do not combine terms. Ka = = 1.5 × 10-5 RESET [0] [0.100] [0.120] [0.140] [0.0200] [0.0400] [x] [2x] [0.100 + x] [0.100-x] [0.120+x] [0.120-x] [0.140 + x] [0.140-x] [0.0200 + x] [0.0200 - x] [0.0400+x] [0.0400 - x] < PREV 2 3 Based on your ICE table and the equilibrium expression for Ka, determine the pH of this solution. pH = RESET 0 4.35 4.49 x 10-5 9.65 3.99 10.0 2.23 x 10-10 1.40 3.23
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter12: Chemical Equilibrium
Section12.6: Shifting A Chemical Equilibrium: Le Chatelier's Principle
Problem 12.10CE
Related questions
Question
Transcribed Image Text:constructing a BCA table, constructing an ICE table, writing the
equilibrium constant expression, and use this information to determine
the pH. The value of Ka for HC-H-O: is 1.5 x 10%. Complete Parts 1-4
before submitting your answer.
NEXT >
40.0 mL of a 0.200 M HC-H-O: solution was titrated with 10.0 mL of 0.100 M Sr(OH): (a
strong base). Fill in the BCA table with the appropriate value for each involved species
to determine the moles of reactant and product after the reaction of the acid and base.
You can ignore the amount of water in the reaction.
HCH.O.(aq)
+
OH(aq)
HO(1)
.
CHO (0)
Before (mol)
Change (mol)
After (mol)
338
RESET
0.100
0.200
1.00 x 10-
-1.00 × 10-³
2.00 x 10-
-2.00 x 10-³
6.00 × 10 -6.00 × 10 7.00 × 10
-7.00 × 10°³ 8.00 × 10
-8.00 × 10
< PREV
NEXT >
Upon completion of the acid-base reaction, the C-H-O₂- ion is in equilibrium with water.
Set up the ICE table in order to determine the unknown concentrations of reactants and
products.
HC.H-Ox(aq)
+
H:O(1)
=
HO-(aq)
+
CH-O:-(aq)
Initial (M)
Change (M)
Equilibrium (M)
RESET
0.100
0.120
0.140
0.0200
0.0400
0.100 + x
0.100 - x
0.120 + x
0.120-x
0.140+x
0.140-x
0.0200 + x
0.0200-x
0.0400+x
0.0400-x
![< PREV
2
3
4
NEXT >
Based on your ICE table and the equilibrium expression for Ka, set up the expression
for Ka in order to determine the unknown concentrations. Each reaction participant
must be represented by one tile. Do not combine terms.
Ka
=
= 1.5 × 10-5
RESET
[0]
[0.100]
[0.120]
[0.140]
[0.0200]
[0.0400]
[x]
[2x]
[0.100 + x]
[0.100-x]
[0.120+x]
[0.120-x]
[0.140 + x]
[0.140-x]
[0.0200 + x]
[0.0200 - x]
[0.0400+x]
[0.0400 - x]
< PREV
2
3
Based on your ICE table and the equilibrium expression for Ka, determine the pH of this
solution.
pH
=
RESET
0
4.35
4.49 x 10-5
9.65
3.99
10.0
2.23 x 10-10
1.40
3.23](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff922ab2d-c754-4c9c-9d3b-bd784fecfd89%2Fd4554ab3-bec6-4699-b615-64ff6bb80934%2F71bsgzh_processed.png&w=3840&q=75)
Transcribed Image Text:< PREV
2
3
4
NEXT >
Based on your ICE table and the equilibrium expression for Ka, set up the expression
for Ka in order to determine the unknown concentrations. Each reaction participant
must be represented by one tile. Do not combine terms.
Ka
=
= 1.5 × 10-5
RESET
[0]
[0.100]
[0.120]
[0.140]
[0.0200]
[0.0400]
[x]
[2x]
[0.100 + x]
[0.100-x]
[0.120+x]
[0.120-x]
[0.140 + x]
[0.140-x]
[0.0200 + x]
[0.0200 - x]
[0.0400+x]
[0.0400 - x]
< PREV
2
3
Based on your ICE table and the equilibrium expression for Ka, determine the pH of this
solution.
pH
=
RESET
0
4.35
4.49 x 10-5
9.65
3.99
10.0
2.23 x 10-10
1.40
3.23
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