One example for NH3 as a conjugate acid: as a conjugate base:
(a) (b) (c) (d) (e) (f)
(a) (b) (c) (d) (e) (f)
(a) H2O, O2−; (b) H3O+, OH−; (c) H2CO3, (d) (e) H2SO4, (f) (g) H2S; S2−; (h) H4N2
The labels are Brønsted-Lowry acid = BA; its conjugate base = CB; Brønsted-Lowry base = BB; its conjugate acid = CA. (a) HNO3(BA), H2O(BB), H3O+(CA), (b) CN−(BB), H2O(BA), HCN(CA), OH−(CB); (c) H2SO4(BA), Cl−(BB), HCl(CA), (d) OH−(BB), (CB), H2O(CA); (e) O2−(BB), H2O(BA) OH−(CB and CA); (f) [Cu(H2O)3(OH)]+(BB), [Al(H2O)6]3+(BA), [Cu(H2O)4]2+(CA), [Al(H2O)5(OH)]2+(CB); (g) H2S(BA), HS−(CB), NH3(CA)
Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2O. As an acid:
As a base:
amphiprotic: (a) (b) not amphiprotic: (c) Br−; (d) (e)
In a neutral solution [H3O+] = [OH−]. At 40 °C,
[H3O+] = [OH−] = (2.910−14)1/2 = 1.7 10−7.
x = 3.051 10−7 M = [H3O+] = [OH−]
pH = −log3.051 10−7 = −(−6.5156) = 6.5156
pOH = pH = 6.5156
(a) pH = 3.587; pOH = 10.413; (b) pH = 0.68; pOH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pH = −0.40; pOH = 14.4
[H3O+] = 3.0 10−7 M; [OH−] = 3.3 10−8 M
[H3O+] = 1 10−2 M; [OH−] = 1 10−12 M
[OH−] = 3.1 10−12 M
The salt ionizes in solution, but the anion slightly reacts with water to form the weak acid. This reaction also forms OH−, which causes the solution to be basic.
[H2O] > [CH3CO2H] > ≈ > [OH−]
The oxidation state of the sulfur in H2SO4 is greater than the oxidation state of the sulfur in H2SO3.
The stronger base or stronger acid is the one with the larger Kb or Ka, respectively. In these two examples, they are (CH3)2NH and
(a) higher electronegativity of the central ion. (b) H2O; NH3 is a base and water is neutral, or decide on the basis of Ka values. (c) HI; PH3 is weaker than HCl; HCl is weaker than HI. Thus, PH3 is weaker than HI. (d) PH3; in binary compounds of hydrogen with nonmetals, the acidity increases for the element lower in a group. (e) HBr; in a period, the acidity increases from left to right; in a group, it increases from top to bottom. Br is to the left and below S, so HBr is the stronger acid.
(a) NaHSeO3 < NaHSO3 < NaHSO4; in polyoxy acids, the more electronegative central element—S, in this case—forms the stronger acid. The larger number of oxygen atoms on the central atom (giving it a higher oxidation state) also creates a greater release of hydrogen atoms, resulting in a stronger acid. As a salt, the acidity increases in the same manner. (b) the basicity of the anions in a series of acids will be the opposite of the acidity in their oxyacids. The acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (c) HOI < HOBr < HOCl; in a series of the same form of oxyacids, the acidity increases as the electronegativity of the central atom increases. Cl is more electronegative than Br, and I is the least electronegative of the three. (d) HOCl < HOClO < HOClO2 < HOClO3; in a series of oxyacids of the same central element, the acidity increases as the number of oxygen atoms increases (or as the oxidation state of the central atom increases). (e) and are anions of weak bases, so they act as strong bases toward H+. and HS− are anions of weak acids, so they have less basic character. In a periodic group, the more electronegative element has the more basic anion. (f) with a larger number of oxygen atoms (that is, as the oxidation state of the central ion increases), the corresponding acid becomes more acidic and the anion consequently less basic.
Strong electrolytes are 100% ionized, and, as long as the component ions are neither weak acids nor weak bases, the ionic species present result from the dissociation of the strong electrolyte. Equilibrium calculations are necessary when one (or more) of the ions is a weak acid or a weak base.
1. Assume that the change in initial concentration of the acid as the equilibrium is established can be neglected, so this concentration can be assumed constant and equal to the initial value of the total acid concentration. 2. Assume we can neglect the contribution of water to the equilibrium concentration of H3O+.
(b) The addition of HCl
(a) Adding HCl will add H3O+ ions, which will then react with the OH− ions, lowering their concentration. The equilibrium will shift to the right, increasing the concentration of HNO2, and decreasing the concentration of ions. (b) Adding HNO2 increases the concentration of HNO2 and shifts the equilibrium to the left, increasing the concentration of ions and decreasing the concentration of OH− ions. (c) Adding NaOH adds OH− ions, which shifts the equilibrium to the left, increasing the concentration of ions and decreasing the concentrations of HNO2. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding KNO2 adds ions and shifts the equilibrium to the right, increasing the HNO2 and OH− ion concentrations.
This is a case in which the solution contains a mixture of acids of different ionization strengths. In solution, the HCO2H exists primarily as HCO2H molecules because the ionization of the weak acid is suppressed by the strong acid. Therefore, the HCO2H contributes a negligible amount of hydronium ions to the solution. The stronger acid, HCl, is the dominant producer of hydronium ions because it is completely ionized. In such a solution, the stronger acid determines the concentration of hydronium ions, and the ionization of the weaker acid is fixed by the [H3O+] produced by the stronger acid.
(a) is the correct statement.
[H3O+] = 7.5 10−3 M
[HNO2] = 0.127
[OH−] = 1.3 10−12 M
[BrO−] = 4.5 10−8 M
[HBrO] = 0.120 M
[OH−] = = 0.0014 M
[NH3] = 0.144 M
[H3O+] = 6.9 10−12 M
= 3.9 10−8 M
[C6H5NH2] = 0.100 M
Solving for x gives 1.63 10−5 M. This value is less than 5% of 0.0092, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [ClO] = 5.8 10−5 M
[HClO] = 0.00092 M
[OH−] = 6.1 10−10 M;
Solving for x gives 5.81 10−6 M. This value is less than 5% of 0.0784, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
= [OH−] = 5.8 10−6 M
[C6H5NH2] = 0.00784
[H3O+] = 1.7 10−9 M;
Solving for x gives 6.30 10−6 M. This value is less than 5% of 0.0810, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[H3O+] = [CN−] = 6.3 10−6 M
[HCN] = 0.0810 M
[OH−] = 1.6 10−9 M;
Solving for x gives 2.63 10−3 M. This value is less than 5% of 0.11, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[(CH3)3NH+] = [OH−] = 2.6 10−3 M
[(CH3)3N] = 0.11 M
[H3O+] = 3.8 10−12 M;
Solving for x gives 1.39 10−4 M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Thus, the concentrations of solute species at equilibrium are:
[Fe(H2O)5(OH)+] = [H3O+] = 1.4 10−4 M
= 0.120 M
[OH−] = 7.2 10−11 M
pH = 2.41
[C10H14N2] = 0.049 M
[C10H14N2H+] = 1.9 10−4 M
= 1.4 10−11 M
[OH−] = 1.9 10−4 M
[H3O+] = 5.3 10−11 M
(a) acidic; (b) basic; (c) acidic; (d) neutral
[H3O+] and are practically equal
[C6H4(CO2H)2] 7.2 10−3 M, [C6H4(CO2H)(CO2)−] = [H3O+] 2.8 10−3 M, 3.9 10−6 M, [OH−] 3.6 10−12 M
Solving for x gives 1.5 10−11 M. Therefore, compared with 0.014 M, this value is negligible (1.1 10−7%).
Excess H3O+ is removed primarily by the reaction:
Excess base is removed by the reaction:
[H3O+] = 1.5 10−4 M
[OH−] = 4.2 10−4 M
[NH4NO3] = 0.36 M
(a) The added HCl will increase the concentration of H3O+ slightly, which will react with and produce CH3CO2H in the process. Thus, decreases and [CH3CO2H] increases.
(b) The added KCH3CO2 will increase the concentration of which will react with H3O+ and produce CH3CO2 H in the process. Thus, [H3O+] decreases slightly and [CH3CO2H] increases.
(c) The added NaCl will have no effect on the concentration of the ions.
(d) The added KOH will produce OH− ions, which will react with the H3O+, thus reducing [H3O+]. Some additional CH3CO2H will dissociate, producing ions in the process. Thus, [CH3CO2H] decreases slightly and increases.
(e) The added CH3CO2H will increase its concentration, causing more of it to dissociate and producing more and H3O+ in the process. Thus, [H3O+] increases slightly and increases.
pH = 8.95
37 g (0.27 mol)
(a) pH = 5.222;
(b) The solution is acidic.
(c) pH = 5.221
To prepare the best buffer for a weak acid HA and its salt, the ratio should be as close to 1 as possible for effective buffer action. The [H3O+] concentration in a buffer of pH 3.1 is [H3O+] = 10−3.1 = 7.94 10−4 M
We can now solve for Ka of the best acid as follows:
In Table 14.2, the acid with the closest Ka to 7.94 10−4 is HF, with a Ka of 7.2 10−4.
For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio that is as close to 1 as possible. The pOH of the buffer is 14.00 − 10.65 = 3.35. Therefore, [OH−] is [OH−] = 10−pOH = 10−3.35 = 4.467 10−4 M.
We can now solve for Kb of the best base as follows:
Kb = [OH−] = 4.47 10−4
In Table 14.3, the base with the closest Kb to 4.47 10−4 is CH3NH2, with a Kb = 4.4 10−4.
The molar mass of sodium saccharide is 205.169 g/mol. Using the abbreviations HA for saccharin and NaA for sodium saccharide the number of moles of NaA in the solution is:
9.75 10−6 mol
The pKa for [HA] is 1.68, so [HA] = M. Thus, [A–] (saccharin ions) is M.
Thus, [A−](saccarin ions) is 3.90 10−5 M
At the equivalence point in the titration of a weak base with a strong acid, the resulting solution is slightly acidic due to the presence of the conjugate acid. Thus, pick an indicator that changes color in the acidic range and brackets the pH at the equivalence point. Methyl orange is a good example.
In an acid solution, the only source of OH− ions is water. We use Kw to calculate the concentration. If the contribution from water was neglected, the concentration of OH− would be zero.
(a) pH = 2.50;
(b) pH = 4.01;
(c) pH = 5.60;
(d) pH = 8.35;
(e) pH = 11.08