Answer
|
0.000256M
=Ka
|
0.061M=
Ka
|
498 M
=Ka
|
0.71%
|
0.00079%
|
0.049M
|
Step 1 Derive the equation for the ionisation.
Step 2 Derive the equilibrium expression.
Step 3 Calculate the value for
each concentration
[hydronium ion] = 0.16 X 2.0 =0.32M
[X ion] = 0.16 X 2.0 =0.32M
[monoprotic acid] = 2.0 - 0.32 M = 1.68M
Step 4 Calculate the Ka value.
0.32 X 0.32 / 1.68 = 0.061M
Step 1 Derive the equation for the ionisation.
Step 2 Derive the equilibrium expression.
Step 3 Calculate the value for
each concentration
[hydronium ion] = 0.016 X 1.0 =0.016M
[acetate ion] = 0.016 X 1.0 =0.016M
[ehtanoic acid] = 1.0M
Step 4 Calculate the Ka value.
Step 1 Derive the equation for the ionisation.
Step 2 Derive the equilibrium expression.
Step 3 Calculate the value for
each concentration
[hydronium ion] = 0.998 X 1.0 =0.998M
[chloride ion] = 0.998 X 1.0 =0.998M
[HCl acid] = 0.002M
Step 4 Calculate the Ka value.
Step 1 Derive the equation for the ionisation.
Step 2 Derive the equilibrium expression.
The concentration of hydronium ion is equal to the
[X ion]
Step 3 Calculate the value for [hydronium ion]
Assume the concentration of the
weak monoprotic acid is unchanged and remains at 2M
[hydronium ion] = 0.0141M
Step 4 Calculate the % ionisation
(0.0141/2) X 100 = 0.71%
Acidity constants Ka |
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