The pH of a solution of 19.5 g of malonic acid in 0.250 L is 1.47.The pH of a 0.300 M Solution of sodium hydrogen malonate is 4.26. What are the values of Ka1and Ka2

Explanation:

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In this case, since the stepwise dissociation of malonic acid which is a diprotic acid that we are going to symbolize by H₂A, is:

The first ionization has the following equilibrium expression:

Whereas the concentration of H⁺ equals the concentration of HA⁻ and is computed via the pH:

Next, we compute the molarity of the 19.5 g of malonic acid (molar mass = 104.06 g/mol) as shown below:

Thus, Ka1 turns out:

Now, for the second ionization, since the 0.300-M sodium hydrogen malonate is the source of HA⁻, and the pH is 4.26, we can compute the concentration of both H⁺ and A⁻² again by considering the pH:

Therefore Ka2 turns out:

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your answers are a,b,,e and f

i hope this

the answers would be

the first one and the second one