Chemistry -acidity constants Ka exercises

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.

(0.016 X 0.016)/1 = 0.000256M

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.

(0.998X 0.998)/0.002 = 498.002M

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
Exercises

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Question	Answer
Find the Ka of a 1.00M solution of ethanoic acid which is 1.60% ionised.	Ka =
Find the Ka of a 2.00M solution of monoprotic acid which is 16.0% ionised.	Ka =
Find the Ka of a 1.00M solution of HCl acid which is 99.8% ionised.	Ka =
Calculate the percent ionisation of a weak monoprotic acid in a 2.00M solution if the ka is 1.00 X 10^-4M	%=
Calculate the percent ionisation of a monoprotic acid in a 2.0M solution if the pH of the solution is 2.8	%=
Find the Ka of a 2.0M solution of an unknown monoprotic acid which is 15.6% ionised.	Ka =

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