What is the ph of 0.25 M tartaric acid? Ka=2.5*10^-5M
Tartatic acid = H2T
........H2T ==> H^+ + HT-
I......0.25......0.....0
C........-x......x.....x
E......0.25-x....x.....x
Substitute the E line into Ka expression and solve for x = (H^+), then convert to pH.
This looks like the wrong Ka for the acid.
To find the pH of a 0.25 M solution of tartaric acid, we first need to calculate the concentration of hydronium ions (H3O+) in the solution.
The formula for tartaric acid is H2C4H4O6.
Tartaric acid is a diprotic acid, meaning it can donate two protons (H+) per molecule.
Given that the concentration of tartaric acid is 0.25 M, and the acid is diprotic, we can assume that after full ionization, the concentration of H3O+ ions will be two times the concentration of tartaric acid.
Thus, the concentration of H3O+ ions is 0.5 M.
Now we need to calculate the pH. The pH is defined as the negative logarithm of the concentration of H3O+ ions:
pH = -log[H3O+]
Using the concentration of H3O+ ions, we can plug in the values to calculate the pH:
pH = -log(0.5)
Using a calculator, we can find that the pH of a 0.25 M tartaric acid solution is approximately 0.301.
To find the pH of a solution of tartaric acid, we can calculate the concentration of hydrogen ions using the given Ka value of 2.5 * 10^-5M.
The Ka value represents the acid dissociation constant, which indicates the extent to which an acid donates a proton (H+) in water. In this case, tartaric acid (H2T) dissociates into hydrogen ions (H+) and the conjugate base (T^-).
The equation for the dissociation of tartaric acid is:
H2T ⇌ H+ + HT-
The Ka expression for this reaction is:
Ka = [H+][HT-] / [H2T]
Given that the initial concentration of tartaric acid is 0.25M, let's assume the concentration of hydrogen ions at equilibrium is x M. Since the ratios are 1:1, the concentration of the conjugate base (HT-) will also be x M.
Thus, the equilibrium expression becomes:
Ka = x * x / (0.25 - x)
Assuming x is much smaller than 0.25, we can simplify the equation:
Ka = x^2 / 0.25
Now, we can rearrange this equation to solve for x:
x^2 = Ka * 0.25
x = √(Ka * 0.25)
Substituting the given Ka value:
x = √(2.5 * 10^-5M * 0.25)
x ≈ 0.0025M
Since we assumed x is much smaller than 0.25, we can use this approximate value to calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.0025)
pH ≈ 2.60
Therefore, the pH of a 0.25M solution of tartaric acid is approximately 2.60.