Question

Calculate the pH of the following solutions: a. $1.2\mathrm{M}{\mathrm{CaBr}}_2$ b. $0.84M{C}_6{H}_{5}N{H}_{3}N{O}_3(K_b\text{ for }{C}_6{H}_{5}N{H}_2=3.8\times {10}^{-10})$ c. $0.57M{\mathrm{KC}}_7{\mathrm{H}}_5{\mathrm{O}}_2(K_{\mathrm{a}}\text{ for }{\mathrm{HC}}_7{\mathrm{H}}_5{\mathrm{O}}_2=6.4\times {10}^{-5})$

Short answer

The pH of the following solutions are: a. 7 b. 4.74 c. 11.52

Step-by-step solution

a. Calculate the concentration of ions in the $CaB{r}_2$ solution

The solution has $1.2M$ $CaB{r}_2$. In this solution, one $C{a}^{2+}$ ion and two $B{r}^{-}$ ions are formed from each molecule of $CaB{r}_2$. As $C{a}^{2+}$ and $B{r}^{-}$ are spectator ions, they do not affect the pH of the solution. Hence, the pH of this solution will be the same as pure water, which is 7.

b.1. Determine the equilibrium expression for the weak base

The given compound is $0.84M$ $C_6{H}_{5}N{H}_{3}N{O}_3$, and the $K_b$ for the weak base $C_6{H}_{5}N{H}_2$ is $3.8\times {10}^{-10}$. In a solution of $C_6{H}_{5}N{H}_{3}N{O}_3$ the cation $C_6{H}_{5}N{H}_3^{+}$ will interact with water molecules and produce $C_6{H}_{5}N{H}_2$ and $H_3{O}^{+}$ ions. So the equilibrium expression for this reaction is: $C_6{H}_{5}N{H}_3^{+}+H_{2}O\rightleftharpoons C_6{H}_{5}N{H}_2+H_3{O}^{+}$

b.2. Set up the ICE table

Let's set up an ICE table (Initial, Change, Equilibrium) for the reaction. For simplicity, denote the $C_6{H}_{5}N{H}_3^{+}$ ion concentration as $B^{+}$. Initial: $B^{+}=0.84M$, $H_3{O}^{+}=0$, $C_6{H}_{5}N{H}_2=0$ Change: $B^{+}=-x$, $H_3{O}^{+}=+x$, $C_6{H}_{5}N{H}_2=+x$ Equilibrium: $B^{+}=0.84-x$, $H_3{O}^{+}=x$, $C_6{H}_{5}N{H}_2=x$ Now, apply the $K_b$ expression: $K_b=\frac{[C_6{H}_{5}N{H}_2][H_3{O}^{+}]}{[B^{+}]}=3.8\times {10}^{-10}$

b.3. Solve for x and calculate pH

Substitute the equilibrium values from the ICE table into the $K_b$ expression: $3.8\times {10}^{-10}=\frac{x^2}{(0.84-x)}$ Because $K_b$ is very small, the value of x will be much smaller than 0.84, so we can simplify the equation and solve for x: $3.8\times {10}^{-10}=\frac{x^2}{0.84}$ $x=\sqrt{3.8\times {10}^{-10}\times 0.84}\approx 1.8\times {10}^{-5}M$ Now we have the concentration of $H_3{O}^{+}$ at equilibrium. Calculate the pH: $pH=-{\log }_{10}([H_3{O}^{+}])=4.74$ The pH of the solution is approximately 4.74.

c.1. Determine the equilibrium expression for the weak acid

In part c, we have a solution of $0.57M$ $K{C}_7{H}_5{O}_2$, with a given $K_a$ for the weak acid $H{C}_7{H}_5{O}_2$, which is $6.4\times {10}^{-5}$. When the salt dissolves, it dissociates into $K^{+}$ ions and $C_7{H}_5{O}_2^{-}$ ions. The $C_7{H}_5{O}_2^{-}$ ion will react with water and produce $H{C}_7{H}_5{O}_2$ and $O{H}^{-}$ ions. The equilibrium expression for this reaction is: $C_7{H}_5{O}_2^{-}+H_{2}O\rightleftharpoons H{C}_7{H}_5{O}_2+O{H}^{-}$

c.2. Set up the ICE table

Set up an ICE table for the reaction. Let $A^{-}$ represent the concentration of the $C_7{H}_5{O}_2^{-}$ ions. Initial: $A^{-}=0.57M$, $O{H}^{-}=0$, $H{C}_7{H}_5{O}_2=0$ Change: $A^{-}=-x$, $O{H}^{-}=+x$, $H{C}_7{H}_5{O}_2=+x$ Equilibrium: $A^{-}=0.57-x$, $O{H}^{-}=x$, $H{C}_7{H}_5{O}_2=x$ Now, apply the $K_a$ expression: $K_a=\frac{[H{C}_7{H}_5{O}_2][O{H}^{-}]}{[A^{-}]}=6.4\times {10}^{-5}$

c.3. Solve for x and calculate pH

Substitute the equilibrium values from the ICE table into the $K_a$ expression: $6.4\times {10}^{-5}=\frac{x^2}{(0.57-x)}$ As the value of x will be much smaller than 0.57, we can approximate the equation and solve for x: $6.4\times {10}^{-5}=\frac{x^2}{0.57}$ $x=\sqrt{6.4\times {10}^{-5}\times 0.57}\approx 0.0033M$ Now we have the concentration of $O{H}^{-}$ at equilibrium. Calculate the pOH and finally the pH: $pOH=-{\log }_{10}([O{H}^{-}])=2.48$ $pH=14-pOH=14-2.48=11.52$ The pH of the solution is approximately 11.52.

Key concepts

pH Calculation

Understanding how to calculate pH is fundamental in chemistry, particularly for identifying the acidity or basicity of a solution. The pH scale ranges from 0 to 14, with lower values representing acidic conditions, higher values indicating basic conditions, and a pH of 7 being neutral, like pure water. The pH is calculated using the concentration of hydrogen ions $[H_3{O}^{+}]$ in the solution, with the pH formula $pH=-lo{g}_{10}([H_3{O}^{+}])$.

In cases involving weak acids or bases, the concentration of $[H_3{O}^{+}]$ or $[O{H}^{-}]$ must be found using equilibrium calculations before the pH can be determined. For bases, the pOH is found first, which is then used to calculate the pH by subtracting the pOH from 14 (since $pH+pOH=14$ in aqueous solutions). This method must be applied thoughtfully, considering any assumptions or approximations made during the process.

Equilibrium Expression

The equilibrium expression defines the ratio of concentrations of products to reactants at equilibrium, with each raised to the power of their respective coefficients in the balanced chemical equation. For acids and bases, this is typically represented as an acid dissociation constant $K_a$ or a base dissociation constant $K_b$.

For instance, in the case of a weak acid $HA$, the equilibrium expression is $K_a=\frac{[A^{-}][H_3{O}^{+}]}{[HA]}$, where $[A^{-}]$ and $[H_3{O}^{+}]$ are the concentrations of the conjugate base and the hydronium ions at equilibrium, respectively, and $[HA]$ is the concentration of the undissociated acid. These constants are essential for predicting the direction and extent of acid-base reactions.

ICE Table Method

The ICE Table method (Initial, Change, Equilibrium) is a systematic approach to solving equilibrium problems in weak acid and base reactions. It allows one to organize and visualize the initial concentrations, the changes that occur as the system moves towards equilibrium, and the final equilibrium concentrations.

The 'Initial' row indicates the starting concentrations of reactants and products, 'Change' row shows the amount by which each concentration will increase or decrease, often represented by a variable $x$, and the 'Equilibrium' row represents the concentrations of each species once equilibrium is reached. Solving for $x$ and then applying the equilibrium constant expression can determine the equilibrium concentrations necessary for calculating pH.

Weak Acid and Base Reactions

Weak acids and bases do not completely dissociate in solution. They establish an equilibrium between the undissociated and dissociated species in water. A weak acid (HA) typically dissociates partially into $H^{+}$ (or $H_3{O}^{+}$ in water) and its conjugate base $A^{-}$, whereas a weak base (B) reacts with water to produce its conjugate acid $B{H}^{+}$ and hydroxide ions $O{H}^{-}$.

These reactions rely on the acid or base dissociation constants, $K_a$ and $K_b$ respectively, which indicate the strength of the acid or base. To understand the pH of solutions of weak acids and bases, one must calculate the extent of dissociation using $K_a$ or $K_b$ and then determine the concentration of the acidic or basic ions in equilibrium, which ultimately affects the solution's pH.