Question

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

Answer 1

Answer:

$Ka_1=1.61x10^{-3}$

$Ka_2=1.01x10^{-8}$

Explanation:

Hello.

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

$H_2A\rightleftharpoons H^++HA^-;Ka_1\\\\HA^-\rightleftharpoons H^++A^-;Ka_2$

The first ionization has the following equilibrium expression:

$Ka_1=\frac{[H^+][HA^-]}{[H_2A]}$

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

$[H^+]=[HA^-]=10^{-pH}=10^{-1.47}=0.0339M$

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

$[H_2A]=\frac{19.5g/(104.06 g/mol)}{0.250L}=0.750M$

Thus, Ka1 turns out:

$Ka_1=\frac{(0.0339)(0.0339)}{0.750-0.0339}=1.61x10^{-3}$

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:

$[H^+]=[A^-^2]=10^{-4.26}=5.50x10^{-5}M$

Therefore Ka2 turns out:

$Ka_2=\frac{[H^+][A^{-2}]}{[HA^-]}=\frac{(5.50x10^{-5})(5.50x10^{-5})}{0.300-(5.50x10^{-5})}\\ \\Ka_2=1.01x10^{-8}$

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