Question

Using data from Appendix \(\mathrm{D}\), calculate \(\left[\mathrm{OH}^{-}\right]\)and \(\mathrm{pH}\) for each of the following solutions: (a) \(0.105 \mathrm{M} \mathrm{NaF}\), (b) \(0.035 \mathrm{MNa}_{2} \mathrm{~S}\), (c) a mixture that is \(0.045 \mathrm{M}\) in \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(0.055 \mathrm{M}\) in \(\mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\).

Short answer

For the given solutions, the concentrations of hydroxide ions and pH values are as follows: (a) For \(0.105 \mathrm{M}\, \mathrm{NaF}\), \([\mathrm{OH}^{-}] = 1.27\times10^{-6}\, \mathrm{M}\) and \( \mathrm{pH} = 8.10\). (b) For \(0.035 \mathrm{M}\, \mathrm{Na}_2 \mathrm{S}\), \([\mathrm{OH}^{-}] = 2.60\times10^{-4}\, \mathrm{M}\) and \(\mathrm{pH} = 10.41\). (c) For a mixture of \(0.045 \mathrm{M}\, \mathrm{NaCH_{3}COO}\) and \(0.055 \mathrm{M}\, \mathrm{Ba(CH_{3}COO)_2}\), \([\mathrm{OH}^{-}] = 2.91\times10^{-5}\, \mathrm{M}\) and \(\mathrm{pH} = 9.46\).

Step-by-step solution

(a) \(0.105 \mathrm{M} \mathrm{NaF}\)

First, we need to recognize that sodium fluoride (\(\mathrm{NaF}\)) will dissociate into ions in water. The chemical equation for this reaction is: \[\mathrm{NaF} \rightarrow \mathrm{Na^{+}} + \mathrm{F^{-}}\] The fluoride ion (\(\mathrm{F}^{-}\)) can react with water as a base, according to the following equation: \[\mathrm{F}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HF} + \mathrm{OH}^{-}\] Next, we need to find the \(K_\mathrm{b}\) for this reaction using the relationship: \[K_{a} \times K_{b} = K_{w}\] Where, \(K_{(a)}\) is the acid dissociation constant for \(\mathrm{HF}\) (found in Appendix D) and \(K_{w}\) is the ion product of water (\(1.0\times10^{-14}\) at 25°C). For \(\mathrm{HF}\), \(K_{a} = 6.6\times10^{-4}\). Now, we can calculate \(K_b\): \[\begin{aligned} K_{b}=\frac{K_{w}}{K_{a}} &=\frac{1.0\times10^{-14}}{6.6\times10^{-4}} \\ &= 1.52\times10^{-11} \end{aligned}\] Using an ICE (Initial, Change, Equilibrium) table and assuming that \(x\) is the amount of \(\mathrm{OH}^{-}\) and \(\mathrm{HF}\) formed, we can determine the equilibrium concentrations: \[\begin{tabular}{l l l l} & Initial & Change & Equilibrium \\ F⁻ & 0.105 & -x & 0.105-x \\ HF & 0 & +x & x \\ OH⁻ & 0 & +x & x \\ \end{tabular}\] Now, we write the equilibrium expression for the base reaction using \(K_b\): \[K_b = \frac{[\mathrm{HF}][\mathrm{OH}^{-}]}{[\mathrm{F}^{-}]} = \frac{(x)(x)}{(0.105-x)}\] Substituting \(K_b\) value and solving for \(x\), we get: \(x^2 = K_b \times (0.105-x)\) As the value of \(K_b\) is very small, we can assume \(x << 0.105\). Therefore, the equation becomes: \[x^2 = K_b \times 0.105\] Now, we can solve for \(x\): \[\begin{aligned} x=\sqrt{1.52\times10^{-11} \times 0.105} &= 1.27\times10^{-6} \end{aligned}\] So, the concentration of hydroxide ion, \([\mathrm{OH}^{-}]\), is \(1.27\times10^{-6}\). To find the pH, we can calculate the concentration of \([\mathrm{H^{+}}]\) using relationship \([\mathrm{H}^{+}][\mathrm{OH}^{-}]=K_{w}\) and then use the definition of \(\mathrm{pH}\): \[\begin{aligned} [\mathrm{H}^{+}]&=\frac{K_{w}}{[\mathrm{OH}^{-}]}\\&=\frac{1\times10^{-14}}{1.27\times10^{-6}}\\ &= 7.87\times10^{-9} \end{aligned}\] Now, we calculate the pH: \[\begin{aligned} \mathrm{pH} &= -\log [\mathrm{H}^{+}]\\ &=-\log(7.87\times10^{-9})\\ &= 8.10 \end{aligned}\] For solution (a), \([\mathrm{OH}^{-}] = 1.27\times10^{-6}\, \mathrm{M}\) and \(\mathrm{pH} = 8.10\).

(b) \(0.035 \mathrm{M} \mathrm{Na}_2 \mathrm{S}\)

Sodium sulfide (\(\mathrm{Na}_2 \mathrm{S}\)) dissociates into ions in water: \[\mathrm{Na}_2 \mathrm{S} \rightarrow 2\mathrm{Na^{+}} + \mathrm{S^{2-}}\] The \(\mathrm{S^{2-}}\) ion can react with water as a base, according to the following equation: \[\mathrm{S^{2-}} + 2\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HS}^- + 2\mathrm{OH}^{-}\] Now, the process to find the concentration of hydroxide ion and pH would be similar to (a). However, this is a polyprotic system where the \(\mathrm{HS}^-\) could also react with water to produce \(\mathrm{H}_2\mathrm{S}\) and more \(\mathrm{OH}^{-}\). In such cases, the contribution of the first deprotonation step to \([\mathrm{OH}^{-}]\) is the dominant contribution. Therefore, we only consider the first reaction. Find the \(K_\mathrm{b1}\) for \(\mathrm{S^{2-}}\) using the relationship: \[K_{a1} \times K_{b1} = K_{w}\] For \(\mathrm{H}_2\mathrm{S}\), \(K_{a1} = 1.0 \times 10^{-7}\). Now, we can calculate \(K_{b1}\): \[\begin{aligned} K_{b1}=\frac{K_{w}}{K_{a1}} &=\frac{1.0\times10^{-14}}{1.0 \times 10^{-7}} \\ &= 1.0\times10^{-7} \end{aligned}\] Using an ICE table: \[\begin{tabular}{l l l l} & Initial & Change & Equilibrium \\ S²⁻ & 0.035 & -x & 0.035-x \\ HS⁻ & 0 & +x & x \\ OH⁻ & 0 & +2x & 2x \\ \end{tabular}\] Now, we write the equilibrium expression for the base reaction using \(K_{b1}\): \[K_{b1} = \frac{[\mathrm{HS^-}][\mathrm{OH}^{-}]^2}{[\mathrm{S^{2-}}]} = \frac{(x)(2x)^2}{0.035-x}\] Substituting \(K_{b1}\) value and solving for \(x\), we get: \[(2x)^2 = K_{b1} \times (0.035-x)\] Again, as the value of \(K_{b1}\) is relatively small, we can assume \(x << 0.035\). Therefore, the equation becomes: \[(2x)^2 = K_{b1} \times 0.035\] Now, we can solve for \(x\): \[\begin{aligned} x = \sqrt{\frac{(1.0\times10^{-7}) \times 0.035}{4}} &= 1.30\times10^{-4} \end{aligned}\] So, the concentration of hydroxide ion, \([\mathrm{OH}^{-}]\), is \(2x = 2.60\times10^{-4} \mathrm{M}\). To find the pH, we can follow the same steps as before: \[\begin{aligned} [\mathrm{H}^{+}]&=\frac{K_{w}}{[\mathrm{OH}^{-}]}\\&=\frac{1\times10^{-14}}{2.60\times10^{-4}}\\ &= 3.85\times10^{-11} \end{aligned}\] Now, we calculate the pH: \[\begin{aligned} \mathrm{pH} &= -\log [\mathrm{H}^{+}]\\ &=-\log(3.85\times10^{-11})\\ &= 10.41 \end{aligned}\] For solution (b), \([\mathrm{OH}^{-}] = 2.60\times10^{-4}\, \mathrm{M}\) and \(\mathrm{pH} = 10.41\).

(c) \(0.045 \mathrm{M} \mathrm{NaCH_{3}COO}\) and \(0.055 \mathrm{M} \mathrm{Ba(CH_{3}COO)_2}\)

In this case, we have two different compounds, both of which are salts of acetate. We can treat both \(\mathrm{NaCH_3COO}\) and \(\mathrm{Ba(CH_3COO)_2}\) as sources of \(\mathrm{CH_3COO^-}\) ions, reacting with water as bases to form \(\mathrm{CH_3COOH}\) and \(\mathrm{OH^-}\) ions. The total concentration of \(\mathrm{CH_3COO^-}\) ions in the solution is \(0.045 \mathrm{M} + 2\times0.055 \mathrm{M} = 0.155 \mathrm{M}\). The equilibrium reaction: \[\mathrm{CH}_3\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} + \mathrm{OH}^{-}\] Use the same method to find \(K_b\) and subsequently the equilibrium concentrations of each species: For \(\mathrm{CH}_3\mathrm{COOH}\), \(K_{a} = 1.8\times10^{-5}\). Now, we can calculate \(K_b\): \[\begin{aligned} K_{b}=\frac{K_{w}}{K_{a}} &=\frac{1.0\times10^{-14}}{1.8 \times 10^{-5}} \\ &= 5.56\times10^{-10} \end{aligned}\] Using an ICE table: \[\begin{tabular}{l l l l} & Initial & Change & Equilibrium \\ CH₃COO⁻ & 0.155 & -x & 0.155-x \\ CH₃COOH & 0 & +x & x \\ OH⁻ & 0 & +x & x \\ \end{tabular}\] Now, we write the equilibrium expression for the base reaction using \(K_b\): \[K_b = \frac{[\mathrm{CH}_3\mathrm{COOH}][\mathrm{OH}^{-}]}{[\mathrm{CH}_3\mathrm{COO}^{-}]} = \frac{(x)(x)}{(0.155-x)}\] Substituting \(K_b\) value and solving for \(x\), we get: \[x^2 = K_b \times (0.155-x)\] As the value of \(K_b\) is very small, we can assume \(x << 0.155\). Therefore, the equation becomes: \[x^2 = K_b \times 0.155\] Now, we can solve for \(x\): \[\begin{aligned} x=\sqrt{5.56\times10^{-10} \times 0.155} &= 2.91\times10^{-5} \end{aligned}\] So, the concentration of hydroxide ion, \([\mathrm{OH}^{-}]\), is \(2.91\times10^{-5}\). To find the pH, we can follow the same steps as before: \[\begin{aligned} [\mathrm{H}^{+}]&=\frac{K_{w}}{[\mathrm{OH}^{-}]}\\&=\frac{1\times10^{-14}}{2.91\times10^{-5}}\\ &= 3.44\times10^{-10} \end{aligned}\] Now, we calculate the pH: \[\begin{aligned} \mathrm{pH} &= -\log [\mathrm{H}^{+}]\\ &=-\log(3.44\times10^{-10})\\ &= 9.46 \end{aligned}\] For solution (c), \([\mathrm{OH}^{-}] = 2.91\times10^{-5}\, \mathrm{M}\) and \(\mathrm{pH} = 9.46\).

Key concepts

Acid-Base Equilibrium

Understanding acid-base equilibrium is crucial for calculating pH and hydroxide ion concentrations in solutions. In these reactions, water is a participant allowing acids and bases to interact with it.
For example, when sodium fluoride (\(\text{NaF}\)) is dissolved in water, it dissociates to form fluoride ions (\(\text{F}^-\)) and sodium ions (\(\text{Na}^+\)). The fluoride ion acts as a base and reacts with water to generate hydroxide ions (\(\text{OH}^-\)) and hydrofluoric acid (\(\text{HF}\)).
This reaction showcases the acid-base equilibrium:

• An acid, such as \(\text{HF}\), donates protons to water.
• The base, like \(\text{F}^-\), accepts a proton from water, forming its conjugate acid, \(\text{HF}\).
• The reversible nature of these reactions ensures that each component finds a balance, arriving at an equilibrium state.

Recognizing these equilibria helps predict the behavior of ions in solutions, informing the calculations for hydroxide ion concentrations and subsequent pH values.

Hydroxide Concentration

Hydroxide ion concentration (\([\text{OH}^-]\)) is pivotal in determining the basicity of a solution. In aqueous solutions, bases dissolve to yield hydroxide ions, while acids supply hydrogen ions (\([\text{H}^+]\)).
Calculating \([\text{OH}^-]\) begins with understanding the base dissociation process.

• Take sodium sulfide (\(\text{Na}_2\text{S}\)), which dissociates to provide sulfide ions (\(\text{S}^{2-}\)).
• The \(\text{S}^{2-}\) ions interact with water to form hydroxide ions, following this equation: \(\text{S}^{2-} + 2\text{H}_2\text{O} \rightleftharpoons \text{HS}^- + 2\text{OH}^-\).
• The concentration at equilibrium relies on the extent of this reaction, influenced by the dissociation constants.

Once \([\text{OH}^-]\) is established, it can be used to determine the pH by balancing both the hydroxide and hydrogen ion concentrations in the relationship \([\text{H}^+][\text{OH}^-] = K_w\), where \(K_w\) is the water ionization constant (\(1.0 \times 10^{-14}\) at \(25^\circ C\)).

Dissociation Constant

Dissociation constants (\(K_a\) and \(K_b\)) quantify the extent to which acids or bases dissociate in solution. These constants help determine the equilibrium concentrations of the different species:

• The acid dissociation constant \(K_a\) refers to the equilibrium constant for the dissociation of acids in water.
• The base dissociation constant \(K_b\) relates to bases and their interactions with water, converting into conjugate acids and hydroxide ions.

Take the acetate ion scenario, where sodium acetate and barium acetate provide acetate ions (\(\text{CH}_3\text{COO}^-\)). The reaction is explained by the equation: \(\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-\).
The \(K_a\) for acetic acid and water's \(K_w\) help us calculate the \(K_b\) using \(K_w = K_a \times K_b\). A small \(K_b\) shows that the base weakly dissociates in water compared to strong bases.

ICE Table Analysis

The ICE (Initial, Change, Equilibrium) table is a powerful analytical tool for understanding equilibrium states in chemical reactions. It helps track concentration changes from initial states to equilibrium:

• Initial: Begin with the initial molar concentrations of all species involved.
• Change: Mark the shifts that occur in concentration as the reaction progresses towards equilibrium.
• Equilibrium: Reflect the final concentrations at equilibrium using algebraic expressions.

Consider the acetate ion \(\text{CH}_3\text{COO}^-\) as it goes through a base reaction. Let's break it down:- The initial concentration is the total molarity of the acetate ions.- The change is represented by \(x\), where each mole of acetate ion produces moles of acetic acid and hydroxide ion.- For equilibrium, plug these into the equilibrium expression derived from the dissociation constant.Using these equations greatly simplifies predicting the behavior and concentrations of ions at equilibrium, enabling accurate \([\text{OH}^-]\) and \(\text{pH}\) calculations.