Determine the resulting pH when 0.040 mol of solid NaOH is added to a 200.0 mL buffer containing 0.100 mol C&HSNH3CI and 0.500 M C&HsNH2. The value of Kb for CoHsNH2 is 4.3 x 10 10 - ( 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. C6HSNH2(aq) + H2O(1) он (аq) + C6HSNH3 (aq) Initial (M) Change (M). Equilibriu m (M).
Determine the resulting pH when 0.040 mol of solid NaOH is added to a 200.0 mL buffer containing 0.100 mol C&HSNH3CI and 0.500 M C&HsNH2. The value of Kb for CoHsNH2 is 4.3 x 10 10 - ( 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. C6HSNH2(aq) + H2O(1) он (аq) + C6HSNH3 (aq) Initial (M) Change (M). Equilibriu m (M).
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
3: Set up the expression for KB in order to determine the unknown.
kB =~~~~~ = 4.3x10^-10
4: 3: Based on the ice table and Alan expression determine the ph of the solution
Transcribed Image Text:Question 4 of 11
Submi
Determine the resulting pH when 0.040 mol of solid NaOH is
added to a 200.0 mL buffer containing 0.100 mol C&HSNH:CI and
0.500 M C&HsNH2. The value of Kb for CoHsNH2 is 4.3 x 10-10 -,
( 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.
C6HSNH2(aq) +
H2O(I)
OH (aq)
+ C6HSNH3*(aq)
Initial (M)
Change
(M)
Equilibriu
m (M).
RESET
0.500
0.100
0.040
0.700
0.300
+x
0.500 + x
0.500 - x
0.100 - x
0.040 + x
0.040 - x
0.700 + x
-x
0.100 + x
0.700 - x
0.300 + x
0.300 - x
Transcribed Image Text:Change
Question 4 of 11
Submit otudy
Determine the resulting pH when 0.040 mol of solid NaOH is
added to a 200.0 mL buffer containing 0.100 mol C&HSNH.CI and
0.500 M C&HSNH2. The value of Kb for CH$NH2 is 4.3 x 10-10
ow
1
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.
now
PE
C6H:NH3 (aq) +
он (аq)
H2O(I)
+ C6HSNH2(aq)
Before
(mol)
Log C
After
|(mol)
RESET
0.040
-0.040
0.500
-0.500
0.100
-0.100
iver
+x
0.040 + x
0.040 - x
0.100 + x
0.100 - x
0.060
-0.060
-x
mole
0.140
-0.140
0.020
-0.020
wher
Racts
af.
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