0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was found to be 0.013 M. Determine the value of Ka for the weak acid.

Chemistry & Chemical Reactivity
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Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria
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.030 moles of a weak acid, HA, was dissolved in 2.0L of water to form a solution. At equilibrium the concentration of HA was found to be .013 M. Determine the value of Ka for the weak acid.

0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a
solution. At equilibrium, the concentration of HA was found to be 0.013 M.
Determine the value of Ka for the weak acid.
Transcribed Image Text:0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was found to be 0.013 M. Determine the value of Ka for the weak acid.
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0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a
solution. At equilibrium, the concentration of HA was found to be 0.013 M.
Determine the value of Ka for the weak acid.
( PREV
2
Based on your ICE table and definition of Ka, set up the expression for Ka and then
evaluate it. Do not combine or simplify terms.
Ka
%3D
%3D
5 RESET
[0]
[2.0]
[0.030]
[0.013]
[0.060]
[0.015]
[0.017]
[0.002]
[0.011]
[x]
[2x]
[0.030 + x]
[0.030 - x]
[0.015 + x]
[0.015 - x]
3 x 10-4
3 x 103
0.2
Transcribed Image Text:Question 19 of 29 Submit 0.030 moles of a weak acid, HA, was dissolved in 2.0 L of water to form a solution. At equilibrium, the concentration of HA was found to be 0.013 M. Determine the value of Ka for the weak acid. ( PREV 2 Based on your ICE table and definition of Ka, set up the expression for Ka and then evaluate it. Do not combine or simplify terms. Ka %3D %3D 5 RESET [0] [2.0] [0.030] [0.013] [0.060] [0.015] [0.017] [0.002] [0.011] [x] [2x] [0.030 + x] [0.030 - x] [0.015 + x] [0.015 - x] 3 x 10-4 3 x 103 0.2
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