Polyprotic Acids Practice Questions

Key Concepts and Summary

An acid that contains more than one ionizable proton is a polyprotic acid. These acids undergo stepwise ionization reactions involving the transfer of single protons. The ionization constants for polyprotic acids decrease with each subsequent step; these decreases typically are large enough to permit simple equilibrium calculations that treat each step separately.

Practice Problems

Which of the following concentrations would be practically equal in a calculation of the equilibrium concentrations in a 0.134-M solution of H2CO3, a diprotic acid: $[H_3O^+]$, $[OH^-]$, $[H_2CO_3]$, $[HCO_3^-]$, $[CO_3^{2-}]$? No calculations are needed to answer this question.

Solution

[H3O+] and $[HCO_3^-]$ are practically equal

Calculate the concentration of each species present in a 0.050-M solution of H2S.

Calculate the concentration of each species present in a 0.010-M solution of phthalic acid, C6H4(CO2H)2.

$$C_6H_4(CO_2H)_2(aq)+H_2O(l)⇌H_3O^+(aq)+C_6H_4(CO_2H)(CO_2)^-(aq)\qquad K_a=1.1×10^{-3}$$ $$C_6H_4(CO_2H)(CO_2)(aq)+H_2O(l)⇌H_3O^+(aq)+C_6H_4(CO_2)_2^{2-}(aq)\qquad K_a=3.9×10^{-6}$$

 

Solution

[C6H4(CO2H)2] $7.2×10^{-3}\;M$, [C6H4(CO2H)(CO2)] = [H3O+] $2.8×10^{-3}\;M$, $[C_6H_4(CO_2)_2^{2-}]$ $3.9×10^{-6}\;M$, $[OH^-]$ $3.6×10^{-12}\;M$

Salicylic acid, HOC6H4CO2H, and its derivatives have been used as pain relievers for a long time. Salicylic acid occurs in small amounts in the leaves, bark, and roots of some vegetation (most notably historically in the bark of the willow tree). Extracts of these plants have been used as medications for centuries. The acid was first isolated in the laboratory in 1838.

(a) Both functional groups of salicylic acid ionize in water, with Ka = $1.0×10^{-3}$ for the—CO2H group and $4.2×10^{-13}$ for the −OH group. What is the pH of a saturated solution of the acid (solubility = 1.8 g/L).

(b) Aspirin was discovered as a result of efforts to produce a derivative of salicylic acid that would not be irritating to the stomach lining. Aspirin is acetylsalicylic acid, CH3CO2C6H4CO2H. The −CO2H functional group is still present, but its acidity is reduced, Ka = $3.0×10^{-4}$. What is the pH of a solution of aspirin with the same concentration as a saturated solution of salicylic acid (See Part a).

The ion HTe is an amphiprotic species; it can act as either an acid or a base.

(a) What is Ka for the acid reaction of HTe with H2O?

(b) What is Kb for the reaction in which HTe functions as a base in water?

(c) Demonstrate whether or not the second ionization of H2Te can be neglected in the calculation of [HTe] in a 0.10 M solution of H2Te.

Solution

(a) $K_a=1.5×10^{-11}$

(b) $K_b=4.3×10^{-12}$

(c)$\frac{[Te^{2-}][H_3O^+]}{HTe^-}=\frac{(x)(0.0141+x)}{(0.0141-x)}≈\frac{(x)(0.0141)}{0.0141}=1.5×10^{-11}$



Solving for x gives $1.5×10^{-11}\;M$. Therefore, compared with 0.014 M, this value is negligible ($1.1×10^{-7}$%).