The equilibrium constant (Ka) for Reaction 3 at 25 °C is 1.80 × 10−4 mol dm−3. Calculate the equilibrium concentration of hydrogen ions if the concentration of formic acid at equilibrium is 0.00500 mol dm−3. (Show all of your working and give your answer to three significant figures, in correct scientific notation.)
Oh yeah! Those values apply to this equation:
HCOOH(aq) <-----> H^+ (aq)+ HCOO^-(aq)
whats the answer
To answer this question, we need to use the equilibrium constant expression for Reaction 3:
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.
Given that the equilibrium constant Ka is 1.80 × 10^−4 mol dm^−3 and the concentration of formic acid [HA] is 0.00500 mol dm^−3, we are asked to calculate the equilibrium concentration of hydrogen ions [H+].
Let's assume x mol dm^−3 is the concentration of hydrogen ions at equilibrium.
Since formic acid (HA) is a weak acid, it dissociates partially into hydrogen ions (H+) and the conjugate base (A-):
HA ⇌ H+ + A-
At equilibrium, the concentration of formic acid (HA) will decrease by x mol dm^−3, and the concentration of hydrogen ions [H+] and the conjugate base [A-] will both increase by x mol dm^−3.
The new concentrations at equilibrium can be expressed as:
[HA] = 0.00500 - x
[H+] = x
[A-] = x
Now, let's substitute these values into the equilibrium constant expression:
Ka = [H+][A-] / [HA]
1.80 × 10^−4 = (x)(x) / (0.00500 - x)
Since Ka is a very small number, we can assume that the concentration of formic acid (HA) will not change significantly compared to its initial concentration. Thus, we can approximately say that 0.00500 - x ≈ 0.00500.
Now, let's solve for x using this approximation:
1.80 × 10^−4 = (x)(x) / 0.00500
Cross-multiplying gives us:
1.80 × 10^−4 * 0.00500 = x^2
Rearranging the equation gives us:
x^2 = (1.80 × 10^−4) * (0.00500)
x^2 = 9.00 × 10^−7
Taking the square root of both sides gives us:
x ≈ 9.49 × 10^−4
Therefore, the equilibrium concentration of hydrogen ions [H+] is approximately 9.49 × 10^−4 mol dm^−3.