Chapter 14

One example for NH₃ as a conjugate acid: ${\text{NH}}_2{}^{\text{−}}+{\text{H}}^{\text{+}}⟶{\text{NH}}_3;$ as a conjugate base: ${\text{NH}}_4{}^{\text{+}}(aq)+{\text{OH}}^{\text{−}}(aq)⟶{\text{NH}}_3(aq)+{\text{H}}_2\text{O}(l)$

(a) ${\text{H}}_3{\text{O}}^{\text{+}}(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{H}}_2\text{O}(l);$ (b) $\text{HCl}(\mathrm{aq})⟶{\text{H}}^{\text{+}}(aq)+{\text{Cl}}^{\text{−}}(aq);$ (c) ${\text{NH}}_3(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{NH}}_2{}^{\text{−}}(aq);$ (d) ${\text{CH}}_3{\text{CO}}_2\text{H}(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}(aq);$ (e) ${\text{NH}}_4{}^{\text{+}}(aq)⟶{\text{H}}^{\text{+}}(aq)+{\text{NH}}_3(aq);$ (f) ${\text{HSO}}_4{}^{\text{−}}(aq)⟶{\text{H}}^{\mathrm{+}}(aq)+{\text{SO}}_4{}^{\text{2−}}(aq)$

(a) ${\text{H}}_2\text{O}(l)+{\text{H}}^{\text{+}}(aq)⟶{\text{H}}_3{\text{O}}^{\text{+}}(aq);$ (b) ${\text{OH}}^{\text{−}}(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{H}}_2\text{O}(l);$ (c) ${\text{NH}}_3(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{NH}}_4{}^{\text{+}}(aq);$ (d) ${\text{CN}}^{\text{−}}(aq)+{\text{H}}^{\text{+}}(aq)⟶\text{HCN}(aq);$ (e) ${\text{S}}^{\mathrm{2-}}(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{HS}}^{\text{−}}(aq);$ (f) ${\text{H}}_2{\text{PO}}_4{}^{\text{−}}(aq)+{\text{H}}^{\text{+}}(aq)⟶{\text{H}}_3{\text{PO}}_4(aq)$

(a) H₂O, O²⁻; (b) H₃O⁺, OH⁻; (c) H₂CO₃, ${\text{CO}}_3{}^{\text{2−}};$ (d) ${\text{NH}}_4{}^{\text{+}},$ ${\text{NH}}_2{}^{\text{−}};$ (e) H₂SO₄, ${\text{SO}}_4{}^{\text{2−}};$ (f) ${\text{H}}_3{\text{O}}_2{}^{\text{+}},$ ${\text{HO}}_2{}^{\text{−}};$ (g) H₂S; S²⁻; (h) ${\text{H}}_6{\text{N}}_2{}^{\mathrm{2+}},$ 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), ${\text{NO}}_3{}^{\text{−}}(\text{CB});$ (b) CN⁻(BB), H₂O(BA), HCN(CA), OH⁻(CB); (c) H₂SO₄(BA), Cl⁻(BB), HCl(CA), ${\text{HSO}}_4{}^{\text{−}}(\text{CB});$ (d) ${\text{HSO}}_4{}^{\text{−}}(\text{BA}),$ OH⁻(BB), ${\text{SO}}_4{}^{\text{2−}}$(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), ${\text{NH}}_2{}^{\text{−}}(\text{BB}),$ 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:
${\text{H}}_2\text{O}(aq)+{\text{NH}}_3(aq)\rightleftharpoons {\text{NH}}_4{}^{\text{+}}(aq)+{\text{OH}}^{\text{−}}(aq).$ As a base: ${\text{H}}_2\text{O}(aq)+\text{HCl}(aq)\rightleftharpoons {\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{Cl}}^{\text{−}}(aq)$

amphiprotic: (a) ${\text{NH}}_3+{\text{H}}_3{\text{O}}^{\text{+}}⟶{\text{NH}}_4\text{OH}+{\text{H}}_2\text{O},$ ${\text{NH}}_3+{\text{OCH}}_3{}^{\text{−}}⟶{\text{NH}}_2{}^{\text{−}}+{\text{CH}}_3\text{OH};$ (b) ${\text{HPO}}_4{}^{\text{2−}}+{\text{OH}}^{\text{−}}⟶{\text{PO}}_4{}^{\text{3−}}+{\text{H}}_2\text{O},$ ${\text{HPO}}_4{}^{\text{2−}}+{\text{HClO}}_4⟶{\text{H}}_2{\text{PO}}_4{}^{\text{−}}+{\text{ClO}}_4{}^{\text{−}};$ not amphiprotic: (c) Br⁻; (d) ${\text{NH}}_4{}^{\text{+}};$ (e) ${\text{AsO}}_4{}^{\text{3−}}$

In a neutral solution [H₃O⁺] = [OH⁻]. At 40 °C,
[H₃O⁺] = [OH⁻] = (2.910⁻¹⁴)^1/2 = 1.7 $\times$ 10⁻⁷.

x = 3.051 $\times$ 10⁻⁷ M = [H₃O⁺] = [OH⁻]
pH = −log3.051 $\times$ 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 $\times$ 10⁻⁷ M; [OH⁻] = 3.3 $\times$ 10⁻⁸ M

[H₃O⁺] = 1 $\times$ 10⁻² M; [OH⁻] = 1 $\times$ 10⁻¹² M

[OH⁻] = 3.1 $\times$ 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] > $[{\text{H}}_3{\text{O}}^{\text{+}}]$ ≈ $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ > [OH⁻]

The oxidation state of the sulfur in H₂SO₄ is greater than the oxidation state of the sulfur in H₂SO₃.

$\begin{array}{cccccc}\text{Mg}{(\text{OH})}_2(s)+& \text{2HCl}(aq)& ⟶& {\text{Mg}}^{\mathrm{2+}}(aq)+& 2{\text{Cl}}^{\text{−}}(aq)+& {\text{2H}}_2\text{O}(l)\\ \text{BB}& \text{BA}& & \text{CB}& \text{CA}& \end{array}$

$K_{\text{a}}=2.3\times {10}^{\mathrm{-11}}$

The stronger base or stronger acid is the one with the larger K_b or K_a, respectively. In these two examples, they are (CH₃)₂NH and ${\text{CH}}_3{\text{NH}}_3{}^{\text{+}}.$

triethylamine.

(a) ${\text{HSO}}_4{}^{\text{−}};$ higher electronegativity of the central ion. (b) H₂O; NH₃ is a base and water is neutral, or decide on the basis of K_a 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) ${\text{ClO}}_2{}^{\text{−}}<{\text{BrO}}_2{}^{\text{−}}<{\text{IO}}_2{}^{\text{−}};$ 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) ${\text{HTe}}^{\text{−}}<{\text{HS}}^{\text{−}}<<{\text{PH}}_2{}^{\text{−}}<{\text{NH}}_2{}^{\text{−}};$ ${\text{PH}}_2{}^{\text{−}}$ and ${\text{NH}}_2{}^{\text{−}}$ are anions of weak bases, so they act as strong bases toward H⁺. ${\text{HTe}}^{\text{−}}$ 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) ${\text{BrO}}_4{}^{\text{−}}<{\text{BrO}}_3{}^{\text{−}}<{\text{BrO}}_2{}^{\text{−}}<{\text{BrO}}^{\text{−}};$ 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.

$[{\text{H}}_2\text{O}]>[{\text{C}}_6{\text{H}}_4\text{OH}({\text{CO}}_2\text{H})]>{\text{[H}}^{\text{+}}\text{]}\text{0}>{\text{[C}}_6{\text{H}}_4\text{OH}{({\text{CO}}_2)}^{\text{−}}]\gg [{\text{C}}_6{\text{H}}_4\text{O}{({\text{CO}}_2\text{H})}^{\text{−}}]>[{\text{OH}}^{\text{−}}]$

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 ${\text{NO}}_2{}^{\text{−}}$ ions. (b) Adding HNO₂ increases the concentration of HNO₂ and shifts the equilibrium to the left, increasing the concentration of ${\text{NO}}_2{}^{\text{−}}$ 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 ${\text{NO}}_2{}^{\text{−}}$ ions and decreasing the concentrations of HNO₂. (d) Adding NaCl has no effect on the concentrations of the ions. (e) Adding KNO₂ adds ${\text{NO}}_2{}^{\text{−}}$ 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) $K_{\text{b}}=1.8\times {10}^{\mathrm{-5}};$
(b) $K_{\text{a}}=4.5\times {10}^{\mathrm{-4}};$
(c) $K_{\text{b}}=7.4\times {10}^{\mathrm{-5}};$
(d) $K_{\text{a}}=5.6\times {10}^{\mathrm{-10}}$

$K_{\text{a}}=1.2\times {10}^{\mathrm{-2}}$

(a) $K_{\text{b}}=4.3\times {10}^{\mathrm{-12}};$
(b) $K_{\text{a}}=1.6\times {10}^{\mathrm{-8}};$
(c) $K_{\text{b}}=5.9\times {10}^{\mathrm{-7}};$
(d) $K_{\text{b}}=4.2\times {10}^{\mathrm{-3}};$
(e) $K_{\text{b}}=2.3\times {10}^{\mathrm{-3}};$
(f) $K_{\text{b}}=6.3\times {10}^{\mathrm{-13}}$

(a) is the correct statement.

[H₃O⁺] = 7.5 $\times$ 10⁻³ M
[HNO₂] = 0.127
[OH⁻] = 1.3 $\times$ 10⁻¹² M
[BrO⁻] = 4.5 $\times$ 10⁻⁸ M
[HBrO] = 0.120 M

[OH⁻] = $[{\text{NO}}_4{}^{\text{+}}]$ = 0.0014 M
[NH₃] = 0.144 M
[H₃O⁺] = 6.9 $\times$ 10⁻¹² M
$[{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}]$ = 3.9 $\times$ 10⁻⁸ M
[C₆H₅NH₂] = 0.100 M

(a) $\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{ClO}}^{\text{−}}]}{[\text{HClO}]}=\frac{(x)(x)}{(0.0092-x)}\approx \frac{(x)(x)}{0.0092}=2.9\times {10}^{\mathrm{-8}}$
Solving for x gives 1.63 $\times$ 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 $\times$ 10⁻⁵ M
[HClO] = 0.00092 M
[OH⁻] = 6.1 $\times$ 10⁻¹⁰ M;
(b) $\frac{[{\text{C}}_6{\text{H}}_5{\text{NH}}_3{}^{\text{+}}][{\text{OH}}^{\text{−}}]}{[{\text{C}}_6{\text{H}}_5{\text{NH}}_2]}=\frac{(x)(x)}{(0.0784-x)}\approx \frac{(x)(x)}{0.0784}=4.3\times {10}^{\mathrm{-10}}$
Solving for x gives 5.81 $\times$ 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:
$[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ = [OH⁻] = 5.8 $\times$ 10⁻⁶ M
[C₆H₅NH₂] = 0.00784
[H₃O⁺] = 1.7$\times$ 10⁻⁹ M;
(c) $\frac{[{\text{H}}_3{\text{O}}^{\text{+}}][{\text{CN}}^{\text{−}}]}{[\text{HCN}]}=\frac{(x)(x)}{(0.0810-x)}\approx \frac{(x)(x)}{0.0810}=4.9\times {10}^{\mathrm{-10}}$
Solving for x gives 6.30 $\times$ 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 $\times$ 10⁻⁶ M
[HCN] = 0.0810 M
[OH⁻] = 1.6 $\times$ 10⁻⁹ M;
(d) $\frac{[{({\text{CH}}_3)}_3{\text{NH}}^{\text{+}}][{\text{OH}}^{\text{−}}]}{[{({\text{CH}}_3)}_3\text{N}]}=\frac{(x)(x)}{(0.11-x)}\approx \frac{(x)(x)}{0.11}=6.3\times {10}^{\mathrm{-5}}$
Solving for x gives 2.63 $\times$ 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 $\times$ 10⁻³ M
[(CH₃)₃N] = 0.11 M
[H₃O⁺] = 3.8 $\times$ 10⁻¹² M;
(e) $\frac{[\text{Fe}{({\text{H}}_2\text{O})}_5({\text{OH})}^{\text{+}}][{\text{H}}_3{\text{O}}^{\text{+}}]}{[\text{Fe}{({\text{H}}_2\text{O})}_6{}^{\mathrm{2+}}]}=\frac{(x)(x)}{(0.120-x)}\approx \frac{(x)(x)}{0.120}=1.6\times {10}^{\mathrm{-7}}$
Solving for x gives 1.39 $\times$ 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 $\times$ 10⁻⁴ M
$[\text{Fe}{({\text{H}}_2\text{O})}_6{}^{\mathrm{2+}}]$ = 0.120 M
[OH⁻] = 7.2 $\times$ 10⁻¹¹ M

pH = 2.41

[C₁₀H₁₄N₂] = 0.049 M
[C₁₀H₁₄N₂H⁺] = 1.9 $\times$ 10⁻⁴ M
$[{\text{C}}_{10}{\text{H}}_{14}{\text{N}}_2{\text{H}}_2{}^{\mathrm{2+}}]$ = 1.4 $\times$ 10⁻¹¹ M
[OH⁻] = 1.9 $\times$ 10⁻⁴ M
[H₃O⁺] = 5.3 $\times$ 10⁻¹¹ M

$K_{\text{a}}=1.2\times {10}^{\mathrm{-2}}$

$K_{\text{b}}=1.77\times {10}^{\mathrm{-5}}$

(a) acidic; (b) basic; (c) acidic; (d) neutral

[H₃O⁺] and $[{\text{HCO}}_3{}^{\text{−}}]$ are practically equal

[C₆H₄(CO₂H)₂] 7.2 $\times$ 10⁻³ M, [C₆H₄(CO₂H)(CO₂)⁻] = [H₃O⁺] 2.8 $\times$ 10⁻³ M, $[{\text{C}}_6{\text{H}}_4{({\text{CO}}_2)}_2{}^{\mathrm{2-}}]$3.9 $\times$ 10⁻⁶ M, [OH⁻] 3.6 $\times$ 10⁻¹² M

(a) $K_{\text{a2}}=1.5\times {10}^{\mathrm{-11}};$
(b) $K_{\text{b}}=4.3\times {10}^{\mathrm{-12}};$
(c) $\frac{[{\text{Te}}^{\mathrm{2-}}][{\text{H}}_3{\text{O}}^{\text{+}}]}{[{\text{HTe}}^{\text{−}}]}=\frac{(x)(0.0141+x)}{(0.0141-x)}\approx \frac{(x)(0.0141)}{0.0141}=1.5\text{×}{10}^{\mathrm{-11}}$
Solving for x gives 1.5 $\times$ 10⁻¹¹ M. Therefore, compared with 0.014 M, this value is negligible (1.1 $\times$ 10⁻⁷%).

Excess H₃O⁺ is removed primarily by the reaction:
${\text{H}}_3{\text{O}}^{\text{+}}(aq)+{\text{H}}_2{\text{PO}}_4{}^{\text{−}}(aq)⟶{\text{H}}_3{\text{PO}}_4(aq)+{\text{H}}_2\text{O}(l)$
Excess base is removed by the reaction:
${\text{OH}}^{\text{−}}(aq)+{\text{H}}_3{\text{PO}}_4(aq)⟶{\text{H}}_2{\text{PO}}_4{}^{\text{−}}(aq)+{\text{H}}_2\text{O}(l)$

[H₃O⁺] = 1.5 $\times$ 10⁻⁴ M

[OH⁻] = 4.2 $\times$ 10⁻⁴ M

[NH₄NO₃] = 0.36 M

(a) The added HCl will increase the concentration of H₃O⁺ slightly, which will react with ${\text{CH}}_3{\text{CO}}_2{}^{\text{−}}$ and produce CH₃CO₂H in the process. Thus, $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ decreases and [CH₃CO₂H] increases.
(b) The added KCH₃CO₂ will increase the concentration of $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ 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 $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ ions in the process. Thus, [CH₃CO₂H] decreases slightly and $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ increases.
(e) The added CH₃CO₂H will increase its concentration, causing more of it to dissociate and producing more $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ and H₃O⁺ in the process. Thus, [H₃O⁺] increases slightly and $[{\text{CH}}_3{\text{CO}}_2{}^{\text{−}}]$ 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 $\frac{[{\text{H}}_3{\text{O}}^{\text{+}}]}{K_{\text{a}}}$ 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 $\times$ 10⁻⁴ M
We can now solve for K_a of the best acid as follows:
$\begin{array}{}\\ \\ \\ \\ \frac{[{\text{H}}_3{\text{O}}^{\text{+}}]}{K_{\text{a}}}=1\\ K_{\text{a}}=\frac{[{\text{H}}_3{\text{O}}^{\text{+}}]}{1}=7.94\times {10}^{\mathrm{-4}}\end{array}$
In Table 14.2, the acid with the closest K_a to 7.94 $\times$ 10⁻⁴ is HF, with a K_a of 7.2 $\times$ 10⁻⁴.

For buffers with pHs > 7, you should use a weak base and its salt. The most effective buffer will have a ratio $\frac{[{\text{OH}}^{\text{−}}]}{K_{\text{b}}}$ 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 $\times$ 10⁻⁴ M.
We can now solve for K_b of the best base as follows:
$\frac{[{\text{OH}}^{\text{−}}]}{K_{\text{b}}}=1$
K_b = [OH⁻] = 4.47 $\times$ 10⁻⁴
In Table 14.3, the base with the closest K_b to 4.47 $\times$ 10⁻⁴ is CH₃NH₂, with a K_b = 4.4 $\times$ 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 $\times$ 10⁻⁶ mol
The pKa for [HA] is 1.68, so [HA] = $6.2\times {19}^{\mathrm{–9}}$ M. Thus, [A–] (saccharin ions) is $3.90\times {10}^{\mathrm{–5}}$ M.
Thus, [A⁻](saccarin ions) is 3.90 $\times$ 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 K_w 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