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Chapter 17: Problem 112

What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(1.10 \mathrm{~atm} ?\) The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\) -atm. The \(\mathrm{CO}_{2}\) is an acidic oxide, reacting with \(\mathrm{H}_{2} \mathrm{O}\) to form \(\mathrm{H}_{2} \mathrm{CO}_{3}\).

Short Answer

Expert verified
The pH of water saturated with $\mathrm{CO}_{2}$ at $25^{\circ} \mathrm{C}$ and $1.10 \mathrm{~atm}$ is approximately 7.

Step by step solution

01

Use Henry's law to find the concentration of CO鈧� dissolved in water.

We are given the partial pressure of CO鈧� (P) and the Henry's law constant (K) at 25掳C. We can use Henry's law to find the concentration of dissolved CO鈧� in water: \[C = KP\] \[C = (3.1 \times 10^{-2} \frac{mol}{L\cdot atm}) (1.1 atm)\] \[C = 3.41 \times 10^{-2} \frac{mol}{L}\]
02

Write the chemical equation for the reaction of CO鈧� with H鈧侽 to form H鈧侰O鈧�.

CO鈧� reacts with water to form carbonic acid (H鈧侰O鈧�): \[CO_{2(g)} + H_{2O} \rightleftharpoons H_{2}CO_{3(aq)}\]
03

Determine the concentration of H鈧侰O鈧�.

Since CO鈧� is in equilibrium with H鈧侰O鈧�, we can assume that the concentration of H鈧侰O鈧� is equal to the concentration of dissolved CO鈧�: \[ [H_{2}CO_{3}] = 3.41 \times 10^{-2}\, M \]
04

Write the ionization equation of H鈧侰O鈧� and find the concentration of H鈧僌鈦�.

Carbonic acid (H鈧侰O鈧�) ionizes in water to form hydronium ions (H鈧僌鈦�) and bicarbonate ions (HCO鈧冣伝): \[H_{2}CO_{3} \rightleftharpoons H_{3}O鈦� + HCO鈧冣伝\] This reaction can be considered as negligible, and the majority of H鈧僌鈦� ions come from the water's autoionization: \[2H_{2}O \rightleftharpoons OH鈦� + H鈧僌鈦篭] As CO鈧� is a weak acid and its concentration is relatively low, we can assume the autoionization of water doesn't change significantly. The concentration of H鈧僌鈦� in pure water at 25掳C is approximately \(1 \times 10^{-7}\, M\). Therefore, the concentration of H鈧僌鈦� in water saturated with CO鈧� would not be significantly different. \[ [H_{3}O鈦篯 \approx 1 \times 10^{-7}\, M\]
05

Determine the pH.

Now, we have the concentration of H鈧僌鈦�. We can use the pH formula to find the pH: \[pH = -\log_{10}[H_{3}O鈦篯\] \[pH = -\log_{10}(1 \times 10^{-7})\] The pH of water saturated with CO鈧� at 25掳C and 1.10 atm is approximately 7.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Henry's Law
Henry's law is fundamental to solving problems involving the solubility of gases in liquids. According to this principle, at a constant temperature, the concentration of a dissolved gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. Presented by the formula
\[C = KP\]
where \(C\) is the concentration of the dissolved gas, \(K\) is the Henry鈥檚 law constant for the gas, and \(P\) is the partial pressure. For this exercise, by applying Henry's law, we found the concentration of \(CO_2\) that dissolved in water when the gas was at a partial pressure of 1.10 atm.
Acidic Oxide Reaction with Water
When an acidic oxide like \(CO_2\) dissolves in water, it reacts to form a corresponding acid. In the case of \(CO_2\), it reacts with water to form carbonic acid \(H_2CO_3\). This reaction is reversible and can be represented as:
\[CO_{2(g)} + H_{2}O(l) \rightleftharpoons H_{2}CO_{3(aq)}\]
Understanding this reaction is essential to predict the pH of the solution, as the formed carbonic acid plays a critical role in the acid-base chemistry of the solution.
Equilibrium Concentration of Carbonic Acid
In a state of equilibrium, the rate of the forward reaction equals the rate of the backward reaction. In this context, we assume that all the \(CO_2\) that has dissolved turns into \(H_2CO_3\), primarily because carbonic acid is a weak acid and does not dissociate completely. Therefore, the equilibrium concentration of \(H_2CO_3\) can be equated to the initial concentration of dissolved \(CO_2\), which simplifies the process of calculating the pH of the saturated solution.
Ionization of Carbonic Acid
The ionization of carbonic acid in solution is pivotal in understanding its contribution to the overall hydrogen ion \(H^+\) concentration. Although \(H_2CO_3\) ionizes to produce \(H_3O^+\) and \(HCO_3^-\), its low concentration and weak acidic nature imply that its contribution to the hydrogen ion concentration is negligible. Here, the autoionization of water, which is the self-ionization where water acts as both acid and base, becomes more significant. However, the presence of \(CO_2\) slightly alters the concentration of hydrogen ions due to additional acidity, but not enough to drastically change the pH from that of pure water.
Autoionization of Water
The autoionization of water refers to the self-ionization process where water molecules dissociate into hydroxide \(OH^-\) and hydronium \(H_3O^+\) ions. The equilibrium constant for this dissociation, known as the ion product of water \(K_w\), remains constant at a given temperature. At 25掳C, \(K_w\) is \(1 \times 10^{-14}\) which means in pure water, the concentrations of \(OH^-\) and \(H_3O^+\) are each \(1 \times 10^{-7} M\). In our exercise, even with \(CO_2\) dissolved, the primary source of \(H_3O^+\) remains the autoionization of water, leading to approximate calculations for the pH.

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Most popular questions from this chapter

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}, 0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\) and the pH measured. (a) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{p} K_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)

The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20} .\) The \(\mathrm{AsO}_{4}{ }^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{p} K_{a 1}=2.22\right.\); \(\left.\mathrm{p} K_{a 2}=6.98 ; \mathrm{p} K_{a 3}=11.50\right) .\) When asked to calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water, a student used the \(K_{s p}\) expression and assumed that \(\left[\mathrm{Mg}^{2+}\right]=1.5\left[\mathrm{AsO}_{4}^{3-}\right]\) Why was this a mistake?

A 20.0-mL sample of \(0.200 \mathrm{M}\) HBr solution is titrated with \(0.200 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) of the solution after the following volumes of base have been added: (a) \(15.0 \mathrm{~mL}\) (b) \(19.9 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL},\) (d) \(20.1 \mathrm{~mL},\) (e) \(35.0 \mathrm{~mL}\)

\(\mathrm{~A} 1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution is slowly added to \(10.0 \mathrm{~mL}\) of a solution that is \(0.20 \mathrm{M}\) in \(\mathrm{Ca}^{2+}\) and \(0.30 \mathrm{M}\) in \(\mathrm{Ag}^{+}\). (a) Which compound will precipitate first: \(\mathrm{CaSO}_{4}\left(K_{s p}=2.4 \times 10^{-5}\right)\) or \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{s p}=1.5 \times 10^{-5}\right) ?(\mathbf{b})\) How much \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) solution must be added to initiate the precipitation?

(a) Explain the difference between solubility and solubilityproduct constant. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).
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