15.9-15.10 Polyprotic Acids – Acid Strength and Molecular Structure

Polyprotic Acids Polyprotic acids: acids that have more than one ionizable proton. These acids ionize in successive steps. For example, diprotic acids such as sulfurous acid, H2SO3, ionizes in two steps, while triprotic acids, such as phosphoric acid, H3PO4, ionizes in three steps. Since the first step results in the greatest amount of ionization, when finding pH of a weak polyprotic acid, we will just use Ka1, the first acid ionization constant (we can treat these as if the first step is the only step that contributes to the H3O+ concentration). However, when finding the pH of a strong polyprotic acid, we must take the hydronium ion concentration in each step into account since as mentioned previously, as the concentration of a strong acid becomes smaller, the percent ionization increases.

Let’s Try a Practice Problem! Find the pH of a 0.050 M H2CO3 solution. (Ka1 from pg. 732 = 4.3x10-7) Now remember, formic acid is a diprotic acid, so the ionization occurs in two steps, but the first step contributes the most hydronium ions. H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq) Ka1 = 4.3x10-7 HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq) Ka2 = 5.6x10-11 [H3O+] [HCO3-] x2 Ka1 = --------------------- = ------------ = 4.3x10-7 x = 1.5x10-4 continued  [H2CO3] 0.050-x [H2CO3] M [H3O+]M [HCO3-] M Initial 0.050 ~0 Change -x +x Equilibrium 0.050 - x x

x = 1. 5x10-4 = [H3O+] is x valid. 1. 5x10-4 -------------- x 100 = 0 x = 1.5x10-4 = [H3O+] is x valid ? 1.5x10-4 -------------- x 100 = 0.3%, yes x is valid 0.050 pH = -log[H3O+] = -log(1.5x10-4) = 3.82

Let’s Try Another!!! In the cases where we may be asked to find the concentration of an anion of a weak polyprotic acid, we have to use the [HX-] (this is the conjugate base in the first step of a polyprotic acid) and also the [H3O+]from the first ionization step (the strongest step). Find the [CO32-] of the 0.050 M carbonic acid (H2CO3) solution in For Practice we did before. As a reminder: H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq) Ka1 = 4.3x10-7 HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq) Ka2 = 5.6x10-11 And the [H3O+], and [HCO3-]calculated from the first step were both = 1.5x10-4 M Okay, so now we have to set up an ICE table for the second ionization step, and use our hydronium ion concentration above as the initial hydronium ion concentration. Continued 

[H3O+] [CO32-] Ka2 = -------------------- [HCO3-] (1. 5x10-4 + x) x 5 [H3O+] [CO32-] Ka2 = -------------------- [HCO3-] (1.5x10-4 + x) x 5.6x10-11 = --------------------- 1.5x10-4 – x x = 5.6x10-11 M= [CO32-] Now, if you look back to the start of the problem, you should notice that the concentration of the carbonate ion = Ka2. This general rule applies to all diprotic acids in which the x is small approximation is valid. [HCO3-] M [H3O+]M [CO32-] M Initial 1.5x10-4 Change -x +x Equilibrium 1.5x10-4 – x 1.5x10-4 + x x

Let’s Try Another Practice Problem! Find the pH and [SO42-] of a 0.0075 M sulfuric acid solution. H2SO4(aq) + H2O(l) H3O+(aq) + HSO4-(aq) strong HSO4-(aq) + H2O(l) H3O+(aq) + SO42- Ka2 = 0.012 Here we can not use the x is small approximation so… [H3O+][SO42-] (0.0075 + x)(x) Ka2 = ------------------- = --------------------- = 0.012 continued  [HSO4-] 0.0075-x [HSO4-] M [H3O+]M [SO42-] M Initial 0.0075 ~0.0075 Change -x +x Equilibrium 0.0075 - x 0.0075 + x x

(0. 0075 + x)(x) 0. 012 = -------------------- 0. 0075-x 9. 0x10-5 - 0 (0.0075 + x)(x) 0.012 = -------------------- 0.0075-x 9.0x10-5 - 0.012x = 0.0075x + x2 x2 + 0.0195 – 9.0x10-5 = 0 Now use the quadratic equation to solve for x. -0.0195 +/- 0.01952 −4(1)(−9.0𝑥10−5) ----------------------------------------------------------------------------------- 2(1) x = 3.85x10-3 M = [SO42-] Now to find pH, we first need to calculate [H3O+] [H3O+] = 0.0075 + x = 0.0075 + 3.85X10-3 = 1.14X10-2 pH = -log[H3O+] = -log (1.14x10-2) = 1.945

Acid Strength and Molecular Geometry Let’s first look at binary acids: one in which H is bonded to Y  (any element) In order to be acidic, hydrogen must be the less electronegative element in the pairing, because when it ionizes, it must ionize as H+. H-C is not acidic because this bond is virtually nonpolar due to the similarity in electronegativity values of these two elements. Bond strength also determines whether an acid is strong or weak. Even though as electronegativity increases so does acidity, the stronger the bond (i.e. H-F), the weaker the acid. If we examine these trends in group 6A and 7A hydrides, you will see that from left to right acidity increases, and from top to bottom acidity increases.

Acid Strength and Molecular Geometry (Continued) Now let’s turn our attention to oxyacids; one that has the general formula: H-O-Y. The more electronegative the Y, the more it weakens the H-O bond, making the acid more acidic. The more oxygen atoms attached to the Y, also increases the acidity.  sulfuric acid, a strong acid!  sulfurous acid, a weak acid!

Let’s Try a Practice Problem! Which proton shown in red is more acidic? O H II I H – C – O – H (b) H – C – O – H I H (a) Is more acidic, because the carbon in the first acid is also bonded to an additional oxygen which is drawing the electron density away from the carbon, which in turn, draws the electron density away from the O-H bond.

15.9-15.10 pg.748 #’s 112, 118, and 120 Study for Quiz Chapter 15