Need help with AP chemistry, specifically Acids and Bases
The acid ionization constant, Ka, for propanoic acid, C2H5COOH, is 1.3 10
-5
.
(a) Write the expression for the acid-dissociation constant, Ka for propanoic acid.
(b) Calculate the hydrogen ion concentration, [H
+
], in a 0.300 M propanoic acid.
(c) Calculate the percentage of propanoic acid molecules that are ionized in 0.300 M propanoic
acid.
(a) The expression for the acid-dissociation constant, Ka, for a weak acid can be written as follows:
Ka = [H+][A-] / [HA]
Where [H+] represents the concentration of hydrogen ions, [A-] represents the concentration of the conjugate base, and [HA] represents the concentration of the acid. In the case of propanoic acid (C2H5COOH), the acid-dissociation reaction can be represented as:
C2H5COOH ⇌ C2H5COO- + H+
Therefore, the expression for the acid-dissociation constant (Ka) for propanoic acid would be:
Ka = [C2H5COO-][H+] / [C2H5COOH]
(b) To calculate the hydrogen ion concentration, [H+], in a 0.300 M propanoic acid, we can use the information provided in the question and the fact that initially, all the propanoic acid is undissociated.
Since propanoic acid is a weak acid and Ka = 1.3 × 10^-5, we can assume that only a small fraction of the acid molecules will dissociate. Let's assume x represents the fraction of propanoic acid that dissociates. In this case, the concentration of the propanoic acid remaining would be (0.300 - x) M. Therefore, the concentration of the hydrogen ions would also be x M.
Using the expression for Ka, we can rewrite it as:
Ka = (x)(x) / (0.300 - x)
Since the value of Ka is given as 1.3 × 10^-5, we can substitute this value into the equation:
1.3 × 10^-5 = (x)(x) / (0.300 - x)
Simplifying and rearranging the equation, we can solve for x:
x^2 = (1.3 × 10^-5)(0.300 - x)
x^2 - (1.3 × 10^-5)x + 1.3 × 10^-5 × 0.300 = 0
This equation can be solved using the quadratic formula or more approximated methods such as graphical or numerical methods. Solving for x will give us the hydrogen ion concentration, [H+].
(c) To calculate the percentage of propanoic acid molecules that are ionized in 0.300 M propanoic acid, we need to calculate the fraction of propanoic acid that dissociates, which we represented as x in part (b).
The percentage of propanoic acid molecules that are ionized can be calculated using the following formula:
Percentage ionization = (x / initial concentration of propanoic acid) × 100
Substituting the value of x and the initial concentration of propanoic acid (0.300 M) into the formula will give us the desired percentage.
Let's call propanoic acid HP.
.........HP ==> H^+ + P^-
initial.0.3M.....0.....0
change...-x.......x.....x
equil....0.3-x....x......x
Ka = (H^+)(P^-)/(HP)
Substitute from the ICE chart above and solve for x = (H^+).
%ion = [(H^+)/(HP)]*100 = [(?H^+)/0.3]*100
Post your work if you get stuck.