What is the pH of a 0.030 M solution of oxalic acid? Ka for oxalic acid is 5.9 × 10−2.
To determine the pH of a solution of oxalic acid, we can use the equilibrium equation for the dissociation of oxalic acid:
H2C2O4 ⇌ H+ + HC2O4-
The Ka expression for oxalic acid is:
Ka = [H+][HC2O4-] / [H2C2O4]
Given that the Ka value for oxalic acid is 5.9 × 10−2 and the concentration of oxalic acid as 0.030 M, we can set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the species at equilibrium.
Let's assume that "x" represents the concentration of H+ and HC2O4- formed. Therefore, the equilibrium concentrations are:
[H2C2O4] = 0.030 M - x
[H+] = x
[HC2O4-] = x
Substituting these values into the Ka expression gives:
5.9 × 10−2 = (x)(x) / (0.030 - x)
Since the Ka value is small, we can approximate the value of x as negligible compared to 0.030 in the denominator.
Solving the equation gives:
5.9 × 10^-2 = x^2 / 0.030
x^2 = (5.9 × 10^-2)(0.030)
x^2 = 1.77 × 10^-3
Taking the square root of both sides:
x = √(1.77 × 10^-3)
x ≈ 0.042
Therefore, the concentration of H+ (and HC2O4-) is approximately 0.042 M.
To find the pH of the solution, we can use the equation:
pH = -log[H+]
pH = -log(0.042)
pH ≈ 1.38
Thus, the pH of the 0.030 M solution of oxalic acid is approximately 1.38.
To find the pH of a solution of oxalic acid, we need to calculate the concentration of H+ ions in the solution. The concentration of H+ ions can be determined by using the dissociation constant, Ka, of oxalic acid.
The dissociation reaction of oxalic acid is as follows:
H2C2O4(aq) ⇌ H+(aq) + HC2O4-(aq)
The dissociation constant, Ka, can be expressed as:
Ka = [H+(aq)][HC2O4-(aq)] / [H2C2O4(aq)]
Since oxalic acid is a diprotic acid, it undergoes two stages of ionization. However, the second-stage ionization is quite small compared to the first-stage ionization. Therefore, we can assume that all the oxalic acid will ionize in the first stage.
Let's assume that the concentration of H+ ions formed is represented by x. At equilibrium, the concentration of undissociated oxalic acid will be (0.030 - x).
Now we can write the expression for Ka and substitute the given values:
Ka = x * (0.030 - x) / 0.030
Given that Ka is 5.9 × 10^(-2), we can rearrange the equation to solve for x:
5.9 × 10^(-2) = x * (0.030 - x) / 0.030
Simplifying the equation, we get:
1.77 × 10^(-3) = x * (0.030 - x)
This is a quadratic equation that we can solve to find the value of x (concentration of H+ ions).
Once we determine the value of x, we can calculate the pH using the formula:
pH = -log[H+]
By following these steps, you should be able to find the pH of the given solution of oxalic acid.
You need to find the H+ concentration first.
https://socratic.org/questions/calculate-the-h-of-a-0-530-m-solution-of-oxalic-acid-ka-5-90-x-10-2-3-sig-figs-u
Then, the pH is easy
pH=-1ogH+