A 0.010 M solution of a weak monoprotic acid has a pH of 3.70. What is the acid-ionization constant, Ka, for this acid?
a. 2.0 ´ 10-4
b. 2.0 ´ 10-5
c. 4.0 ´ 10-6
d. 4.0 ´ 10-7
e. 4.0 ´ 10-8
HA ==> H^+ +A^-
Ka = (H^+)(A^-)/(HA)
You know H^+ (from pH), A^- is the same, and you know HA. Solve for Ka.
To find the acid-ionization constant, Ka, for this acid, we can use the pH value and the concentration of the acid solution.
Step 1: Write the balanced chemical equation for the ionization of the weak acid:
HA ⇌ H+ + A-
Step 2: Write the expression for the acid-ionization constant, Ka:
Ka = [H+][A-] / [HA]
Step 3: Since the acid is weak and does not ionize completely, we can assume that the concentration of [H+] and [A-] will be equal to x, and the concentration of [HA] will be 0.01 - x (where x is the extent of ionization).
Step 4: Convert the given pH value to [H+] concentration using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-3.70) = 2.0 × 10^(-4)
Step 5: Substitute the concentrations into the acid-ionization constant expression:
Ka = (2.0 × 10^(-4))(2.0 × 10^(-4)) / (0.01 - x)
Step 6: Since the concentration of x is expected to be small compared to 0.01, we can assume that 0.01 - x ≈ 0.01.
Ka = (2.0 × 10^(-4))(2.0 × 10^(-4)) / 0.01
Ka = 4.0 × 10^(-8) / 0.01
Ka = 4.0 × 10^(-6)
Therefore, the acid-ionization constant, Ka, for this acid is c. 4.0 × 10^(-6).
To find the acid-ionization constant, Ka, for this acid, we can use the pH of the solution. The pH of a solution is related to the concentration of hydrogen ions, H+, in the solution. In this case, we are given that the pH is 3.70.
The pH is defined as the negative logarithm (base 10) of the concentration of hydrogen ions, so we can write:
pH = -log[H+]
To find [H+], we need to convert the pH to a concentration. We can do this by using the inverse of the logarithm, which is 10 raised to the power of the negative pH:
[H+] = 10^(-pH)
Substituting the given pH value of 3.70 into the equation:
[H+] = 10^(-3.70)
Calculating this expression, we find that [H+] is approximately 1.99 x 10^(-4) M.
Remember that the concentration of hydrogen ions in a weak acid solution is the same as the concentration of the acid that has ionized. So, in this case, the concentration of the acid that has ionized is also approximately 1.99 x 10^(-4) M.
Since the initial concentration of the weak acid is given as 0.010 M, we can assume that only a small fraction of the acid has ionized. Therefore, the concentration of the non-ionized acid can be approximated as:
[HA] ≈ initial concentration of acid - concentration of ionized acid
[HA] ≈ 0.010 M - 1.99 x 10^(-4) M
[HA] ≈ 0.0098 M
Now, we can set up the equation for the ionization of the acid:
HA ⇌ H+ + A-
The equilibrium expression for the ionization of a weak acid is given by:
Ka = [H+][A-] / [HA]
Substituting the values we have calculated:
Ka = (1.99 x 10^(-4) M)(1.99 x 10^(-4) M) / 0.0098 M
Simplifying this expression, we find that Ka is approximately 4.04 x 10^(-6).
Therefore, the acid-ionization constant, Ka, for this acid is approximately 4.0 ´ 10^(-6).
So, the correct answer is c. 4.0 ´ 10^(-6).