Malic acid is a weak diprotic organic acid with Ka 1=4.0 ×10^-4 and Ka 2=9.0 ×10^-6 a. Letting t

Malic acid is a weak diprotic organic acid with Ka 1=4.0...

Malic acid is a weak diprotic organic acid with $K_{a 1}=4.0 \times 10^{-4}$ and $K_{a 2}=9.0 \times 10^{-6}$ a. Letting the symbol $\mathrm{H}_{2}$ A represent malic acid, write the chemical equations that represent $K_{a 1}$ and $K_{a 2}$. Write the chemical equation that represents $K_{a 1} \times K_{a 2}$ b. Qualitatively describe the relative concentrations of $\mathrm{H}_{2} \mathrm{~A}, \mathrm{HA}^{-}, \mathrm{A}^{2-},$ and $\mathrm{H}_{3} \mathrm{O}^{+}$ in a solution that is about one molar in malic acid. c. Calculate the $\mathrm{pH}$ of a $0.0175 \mathrm{M}$ malic acid solution and the equilibrium concentration of $\left[\mathrm{H}_{2} \mathrm{~A}\right]$ d. What is the $\mathrm{A}^{2-}$ concentrationin in solutions $\mathrm{D}$ and $\mathrm{c}$ ?

Malic acid is a weak diprotic organic acid with $K_{a 1}=4.0 \times 10^{-4}$ and $K_{a 2}=9.0 \times 10^{-6}$
a. Letting the symbol $\mathrm{H}_{2}$ A represent malic acid, write the chemical equations that represent $K_{a 1}$ and $K_{a 2}$. Write the chemical equation that represents $K_{a 1} \times K_{a 2}$
b. Qualitatively describe the relative concentrations of $\mathrm{H}_{2} \mathrm{~A}, \mathrm{HA}^{-}, \mathrm{A}^{2-},$ and $\mathrm{H}_{3} \mathrm{O}^{+}$ in a solution that is about
one molar in malic acid.
c. Calculate the $\mathrm{pH}$ of a $0.0175 \mathrm{M}$ malic acid solution and the equilibrium concentration of $\left[\mathrm{H}_{2} \mathrm{~A}\right]$
d. What is the $\mathrm{A}^{2-}$ concentrationin in solutions $\mathrm{D}$ and $\mathrm{c}$ ?

Key Concepts

Product of Equilibrium Constants

In sequential reactions, such as the stepwise ionizations of a diprotic acid, the overall equilibrium constant for the reaction converting the original acid to the fully deprotonated form is the product of the individual constants (Ka1 × Ka2). This concept simplifies the analysis of multi-step equilibria by linking them into one overall reaction.

pH Calculation for Weak Acids

Calculating pH of a weak acid solution involves setting up the equilibrium expression using the Ka value and the initial concentration of the acid. Due to partial ionization, approximations or solving a quadratic equation may be necessary to determine the hydrogen ion concentration, which then gives the pH value.

Chemical Equilibrium in Acid-Base Reactions

At chemical equilibrium, the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. In acid-base reactions, the equilibrium expression relates the concentrations of the acid, its conjugate base, and water (or hydronium ion), and is essential for calculating the concentrations of various species present.

Stepwise Ionization of Diprotic Acids

Diprotic acids can donate two protons, but these processes occur in steps with separate equilibrium constants. The first dissociation (Ka1) typically occurs to a greater extent than the second (Ka2). Understanding their stepwise nature is crucial to analyze speciation and the behavior of each form in solution.

Acid Dissociation Constant (Ka)

This constant quantitatively describes the extent to which an acid donates its proton(s) in aqueous solution. It is defined by the equilibrium expression of the acid dissociation reaction and indicates the strength of the acid—the higher the Ka, the stronger the acid, and vice versa.

Transcript

00:01 When calculating the ph of a solution that contains a polyprotic acid, such as the diprotic acid, malic acid, we just focus on ka1.

00:11 Ka1 is usually significantly larger than ka2, such that it is the major contributor to the hydronium ion concentration.

00:20 For this diprotic acid, the ka equilibrium is h2a, reacted with water producing hydronium and ha -1 -h -a -1.

00:29 K2 is when h .a.

00:31 Minus gives up its hydrogen to a second water, producing hydronium and a2 minus.

00:39 Summing these two together, we get h2a plus two waters, go to two hydroniums plus a2 minus, and then this corresponds to a k value that's a product of these two k values.

00:55 So qualitatively described relative concentrations of everything? well, because this is a weak acid, we'll have the greater amount of it and then the concentration of hydronium and h .a.

01:08 Minus will be about the same but the hydronium will be a little bit more because we do get a little bit more with the second equilibrium and also from auto -protelices.

01:21 So although they're about the same, hydrogenium is a little bit more than the h .a.

01:26 Minus and then the h .a.

01:27 Minus is greater than the a2 minus.

01:30 So to calculate ph we'll just focus on k -a -1, and normally what we can do is ph will be equal to the negative log of the hydrogenium concentration, the hydrogen concentration being the square root of k -a, in this case k -a -1, multiplied by the concentration of malic acid...

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