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Calculate ph of a 0.150m solution of malonic acid. Ionization constant for the acid are K(1)= 1.40 x 10 ^ -3 ; K(2) = 2.2 x 10^-6
Accepted Answer
The pH of a 0.150 M solution of malonic acid can be calculated by considering both ionization steps. Malonic acid is a diprotic acid, meaning it can donate two protons (H+).
Here's how to approach the calculation:
1. First ionization:
Malonic acid (H2C3H2O4) ⇌ HC3H2O4- + H+
Set up an ICE table and use the given Ka1 to solve for the concentration of H+ ions from the first ionization.
2. Second ionization:
HC3H2O4- ⇌ C3H2O4^2- + H+
Use the initial concentration of HC3H2O4- from the first ionization, along with the Ka2 value, to calculate the additional H+ ions contributed by the second ionization.
3. Total H+ concentration:
Sum the H+ concentrations from both ionization steps.
4. Calculate pH:
pH = -log[H+]
Note: Since the Ka2 value is much smaller than Ka1, the contribution of H+ from the second ionization might be negligible, but it's still important to consider it for a more accurate pH calculation.
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