Determine the resulting pH when 0.0015 of solid Ba(OH)2 is added to a 0.350 L but containing 0.110 M weak acid, HA, and 0.2 M of its conjugate base, A. The value of K HA is 3.2 x 10-º. 1 3 4 Use the table below to determine the mol reactant and product after the reaction of acid and base. You can ignore the amoun

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Determine the resulting pH when 0.0015 mol
of solid Ba(OH)2 is added to a 0.350 L buffer
containing 0.110M weak acid, HA, and 0.220
M of its conjugate base, A. The value of Ka for
HA is 3.2 x 10-9.
1
2
3
4
Based on the result of the acid-base reaction,
set up the ICE table in order to determine the
unknown.
HA(aq) + H2O(1) =H;0*(aq) + A´(aq)
Initial (M)
Change (M)
Equilibrium
(M)
5 RESET
0.110
0.220
0.101
0.229
0.106
+x
0.110 + x
0.110 - x
0.220 + x
-X
0.220 - x
0.101 + x
0.101 - x
0.229 + x
0.229 - x
0.106 + x
0.106 - x
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Transcribed Image Text:Determine the resulting pH when 0.0015 mol of solid Ba(OH)2 is added to a 0.350 L buffer containing 0.110M weak acid, HA, and 0.220 M of its conjugate base, A. The value of Ka for HA is 3.2 x 10-9. 1 2 3 4 Based on the result of the acid-base reaction, set up the ICE table in order to determine the unknown. HA(aq) + H2O(1) =H;0*(aq) + A´(aq) Initial (M) Change (M) Equilibrium (M) 5 RESET 0.110 0.220 0.101 0.229 0.106 +x 0.110 + x 0.110 - x 0.220 + x -X 0.220 - x 0.101 + x 0.101 - x 0.229 + x 0.229 - x 0.106 + x 0.106 - x Tap here or pull up for additional resources
Determine the resulting pH when 0.0015 mol
of solid Ba(OH)2 is added to a 0.350 L buffer
containing 0.110M weak acid, HA, and 0.220
M of its conjugate base, A. The value of Ka for
HA is 3.2 x 10-9.
1
2
4
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.
HA(aq) + OH (aq) → H20(1) + A(aq)
Before (mol)
Change (mol)
After (mol)
5 RESET
0.110
0.220
0.0015
-0.0015
0.0030
-0.0030
0.0385
0.0770
0.0370
0.0355
0.0750
0.0800
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Transcribed Image Text:Determine the resulting pH when 0.0015 mol of solid Ba(OH)2 is added to a 0.350 L buffer containing 0.110M weak acid, HA, and 0.220 M of its conjugate base, A. The value of Ka for HA is 3.2 x 10-9. 1 2 4 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. HA(aq) + OH (aq) → H20(1) + A(aq) Before (mol) Change (mol) After (mol) 5 RESET 0.110 0.220 0.0015 -0.0015 0.0030 -0.0030 0.0385 0.0770 0.0370 0.0355 0.0750 0.0800 Tap here or pull up for additional resources
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