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What is the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of water saturated with \(\mathrm{CO}_{2}\) at a partial pressure of \(1.10\) atm? The Henry's law constant for \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{~mol} / \mathrm{L}-\mathrm{atm}\). 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 CO2 at a partial pressure of 1.10 atm at 25°C is -\(\log(x)\), where \(x = \sqrt{(4.45 \times 10^{-7})(3.41 \times 10^{-2})}\).

Step by step solution

01

Calculate the concentration of CO2 in the solution

Use Henry's Law to find the concentration of CO2 dissolved in water at the given partial pressure and temperature. The formula for Henry's Law is: \[ C = k_H P \] where \(C\) is the concentration of CO2, \(k_H\) is the Henry's law constant, and \(P\) is the partial pressure of CO2. In our case, \(k_H = 3.1 \times 10^{-2} \frac{\mathrm{mol}}{\mathrm{L}\cdot \mathrm{atm}}\) and \(P = 1.10 \ \mathrm{atm}\). Plugging in the given values, we get: \[ C = (3.1 \times 10^{-2} \ \frac{\mathrm{mol}}{\mathrm{L}\cdot \mathrm{atm}}) (1.10 \ \mathrm{atm}) \]
02

Calculate the concentration of H2CO3

The concentration of CO2 and H2CO3 in the solution will be the same, as CO2 reacts with water to form H2CO3. Therefore, \[ [\mathrm{H}_2\mathrm{CO}_3] = [\mathrm{CO}_2] = (3.1 \times 10^{-2}) (1.10) = 3.41 \times 10^{-2}\ \mathrm{M} \]
03

Set up the ionization expression of H2CO3

The ionization of H2CO3 in water can be represented as: \[ \mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons\ \mathrm{H}^+ + \mathrm{H}\mathrm{CO}_3^- \] The ionization constant, \(K_a\), can be calculated as follows: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{H}\mathrm{CO}_3^-]}{[\mathrm{H}_2\mathrm{CO}_3]} \]
04

Use approximation and calculate the concentration of H+ and HCO3- ions

Since H2CO3 is a weak acid, we can make the approximation that the concentrations of H+ and HCO3- ions are equal. \[ [\mathrm{H}^+] = [\mathrm{H}\mathrm{CO}_3^-] = x \] Taking this approximation, the ionization constant expression becomes: \[ K_a = \frac{(x)(x)}{[\mathrm{H}_2\mathrm{CO}_3]-x} \] Given the \(K_{a1}\) value for H2CO3 is \(4.45 \times 10^{-7}\), we can plug it into the equation. Since the acid is weak, \(x\) will be negligible compared to \([\mathrm{H}_2\mathrm{CO}_3]\), so we can simplify the expression: \[ 4.45 \times 10^{-7} = \frac{x^2}{3.41 \times 10^{-2}} \]
05

Solve for x (the concentration of H+ ions)

Solve for \(x\) using the simplified expression: \[ x^2 = (4.45 \times 10^{-7})(3.41 \times 10^{-2}) \] \[ x = \sqrt{(4.45 \times 10^{-7})(3.41 \times 10^{-2})} \]
06

Calculate pH

Finally, calculate the pH of the solution using the formula: \[ \text{pH} = -\log ([\mathrm{H}^+]) \] Calculating the pH using the H+ ion concentration derived in step 5, we get: \[ \text{pH} = -\log(x) \] Based on these steps and calculations, we can find the pH of water saturated with CO2 at a partial pressure of 1.10 atm at 25°C.

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Key Concepts

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

Understanding Henry's Law in Chemistry
Henry's Law is vital to comprehend when predicting the solubility of gases in liquids, which directly impacts pH calculation in water containing gases like CO2. According to this law, the concentration of a gas dissolved in a liquid is directly proportional to its partial pressure above the liquid. Mathematically, it's expressed as:

\[ C = k_H P \]

Here, \(C\) represents the concentration of the dissolved gas, \(k_H\) is the Henry's law constant specific to each gas and solvent combination at a given temperature, and \(P\) is the partial pressure of the gas. In the context of calculating pH for CO2-saturated water, we use Henry's Law to find the dissolved concentration of CO2, key to determining the acidity of the solution. If students remember this proportionality, it can simplify the beginning steps of various chemistry problems related to gas solubility.
Acidic Oxides and Their Role in Acidity
Acidic oxides, such as CO2, are oxides that react with water to produce acids. When CO2 dissolves in water, it forms carbonic acid (H2CO3), a process central to understanding the carbonate system in chemistry:

\[ \text{CO}_2 (g) + \text{H}_2\text{O} (l) \rightleftharpoons \text{H}_2\text{CO}_3 (aq) \]

The reaction with water is what allows CO2 to influence the pH of the solution. This characteristic is crucial in environmental chemistry, where CO2 in rainwater forms weak carbonic acid, leading to natural water acidity, and importantly, is pivotal in problems like the one at hand, where we need to trace the acidic nature of a dissolved gas in water to its equilibrium chemistry.
Ionization Constant: A Measure of Acid Strength
The ionization constant (K_a) is an essential concept in acids' behavior in aqueous solutions. It quantifies the strength of an acid by describing the degree of ionization into H+ ions and its conjugate base. For a generic weak acid HA, the ionization in water can be expressed as:

\[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \]

And the corresponding ionization constant is:

\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]

A high Ka value indicates a strong acid that ionizes more completely, while a low Ka suggests a weak acid. By understanding the ionization constant, students can better predict and calculate the pH of acidic solutions, as the K_a gives us a direct path to the concentrations of ions in equilibrium.
The Weak Acid Approximation in pH Calculations
The weak acid approximation simplifies acidity calculations by assuming that the ionization of a weak acid does not significantly change its initial concentration. This is useful when the Ka is much smaller than the initial concentration of the acid.

In our exercise, by stating that \( [\text{H}^+] = [\text{HCO}_3^-] = x \), and recognizing that \([\text{H}_2\text{CO}_3]\) minus a tiny \(x\) remains practically unchanged, we simplify the equilibrium expression to:

\[ K_a = \frac{x^2}{[\text{H}_2\text{CO}_3]} \]

Then, solving for \(x\), which represents the concentration of hydrogen ions \([H^+]\), becomes much easier. This approximation is a powerful tool in acid-base chemistry, allowing us to quickly estimate the pH of solutions without complex calculations or assumptions.

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Most popular questions from this chapter

Derive an equation similar to the HendersonHasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

Use information from Appendix \(\mathrm{D}\) to calculate the \(\mathrm{pH}\) of (a) a solution that is \(0.150 \mathrm{M}\) in sodium formate \((\mathrm{HCOONa})\) and \(0.200 \mathrm{M}\) in formic acid \((\mathrm{HCOOH})\) (b) a solution that is \(0.210 \mathrm{M}\) in pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\) and \(0.350 \mathrm{M}\) in pyridinium chloride \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHCl}\right) ;\) (c) a solution that is made by combining \(125 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) hydrofluoric acid with \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) sodium fluoride.

Suppose you want to do a physiological experiment that calls for a \(\mathrm{pH} 6.5\) buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 M\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH} 6.50 ?\) (lgnore any volume change.)

(a) Precipitation of the group 4 cations (Figure 17.22) requires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipitated in group 2 and those precipitated in group \(3 ?\) (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation.

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 \(\mathrm{H}_{3} \mathrm{AsO}_{4}\) \(\left(\mathrm{pK}_{a 1}=2.22 ; \mathrm{pK}_{a 2}=6.98 ; \mathrm{pK}_{a 3}=11.50\right) .\) 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?

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