Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HCIO and 0.37 M NaCIO. The value of Ka for HCIO is 2.9 x 10 8. 2 3 NEXT > Use the table below to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of liquid water in the reaction. HCIO(aq) он (аq) H:O(I) CIO (aq) Before (mol) Change (mol) After (mol) 5 RESET 0.13 0.37 +x 0.13 + x 0.13 - x 0.37 + x 0.37 - x 0.003 -0.003 0.013 -0.013 0.037 -0.037 0.010 -0.010 0.040 -0.040
Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HCIO and 0.37 M NaCIO. The value of Ka for HCIO is 2.9 x 10 8. 2 3 NEXT > Use the table below to determine the moles of reactant and product after the reaction of the acid and base. You can ignore the amount of liquid water in the reaction. HCIO(aq) он (аq) H:O(I) CIO (aq) Before (mol) Change (mol) After (mol) 5 RESET 0.13 0.37 +x 0.13 + x 0.13 - x 0.37 + x 0.37 - x 0.003 -0.003 0.013 -0.013 0.037 -0.037 0.010 -0.010 0.040 -0.040
Chemistry
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ISBN:9781305957404
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Chapter1: Chemical Foundations
Section: Chapter Questions
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Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL buffer containing 0.13 M HClO and 0.37 M NaClO. The value of Ka for HClO is 2.9 × 10⁻⁸
Please solve for the pH
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Question 9 of 15
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Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL
buffer containing 0.13 M HCIO and 0.37 M NaCIO. The value of Ka for HCIO is 2.9 x
10 8.
3
4
NEXT
>
Use the table below to determine the moles of reactant and product after the reaction of the acid and
base. You can ignore the amount of liquid water in the reaction.
HCIO(aq)
ОН (аq)
H2O(1)
CIO (aq)
Before (mol)
Change (mol)
After (mol)
5 RESET
0.13
0.37
+x
0.13 + x
0.13 -
-X
0.37 + x
0.37 - x
0.003
-0.003
0.013
-0.013
0.037
-0.037
0.010
-0.010
0.040
-0.040
+
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Transcribed Image Text:Bb Experiment #6 - Standardization X
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Question 9 of 15
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Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL
buffer containing 0.13 M HCIO and 0.37 M NaCIO. The value of Ka for HCIO is 2.9 x
10 8.
3
4
NEXT
>
Use the table below to determine the moles of reactant and product after the reaction of the acid and
base. You can ignore the amount of liquid water in the reaction.
HCIO(aq)
ОН (аq)
H2O(1)
CIO (aq)
Before (mol)
Change (mol)
After (mol)
5 RESET
0.13
0.37
+x
0.13 + x
0.13 -
-X
0.37 + x
0.37 - x
0.003
-0.003
0.013
-0.013
0.037
-0.037
0.010
-0.010
0.040
-0.040
+
7:35 PM
Type here to search
ENG
T
97
3/25/2021
18
+
II
![Bb Experiment #6 - Standardization X
101 Chem101
b My Questions | bartleby
X +
app.101edu.co
M
Question 9 of 15
Submit
Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL
buffer containing 0.13 M HCIO and 0.37 M NaCIO. The value of Ka for HCIO is 2.9 x
10 8.
PREV
1
3
4
NEXT
>
Based on the result of the acid-base reaction, set up the ICE table in order to determine the unknown.
HCIO(aq)
H2O(1)
H;O*(aq)
CIO (aq)
+
+
Initial (M)
Change (M)
Equilibrium (M)
5 RESET
0.13
0.37
0.010
0.040
0.10
0.40
+x
0.13 + x
0.13- х
0.010 + x
0.010 - x
0.040 + x
0.040 - x
-X
0.10 + x
0.10 - x
0.40 + x
0.40 - x
+
7:35 PM
Type here to search
ENG
T
97
3/25/2021
18
1L
II](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faff04ec1-7866-4a6d-a44e-ba0c0ded0b1b%2Fe3a583ce-3a8f-4d4f-8abd-6010c406b86e%2Fi7jqc4k_processed.png&w=3840&q=75)
Transcribed Image Text:Bb Experiment #6 - Standardization X
101 Chem101
b My Questions | bartleby
X +
app.101edu.co
M
Question 9 of 15
Submit
Determine the resulting pH when 0.003 mol of solid NaOH is added to a 100.0 mL
buffer containing 0.13 M HCIO and 0.37 M NaCIO. The value of Ka for HCIO is 2.9 x
10 8.
PREV
1
3
4
NEXT
>
Based on the result of the acid-base reaction, set up the ICE table in order to determine the unknown.
HCIO(aq)
H2O(1)
H;O*(aq)
CIO (aq)
+
+
Initial (M)
Change (M)
Equilibrium (M)
5 RESET
0.13
0.37
0.010
0.040
0.10
0.40
+x
0.13 + x
0.13- х
0.010 + x
0.010 - x
0.040 + x
0.040 - x
-X
0.10 + x
0.10 - x
0.40 + x
0.40 - x
+
7:35 PM
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ENG
T
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3/25/2021
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1L
II
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