Textbook Question
Calculate the pH and the concentrations of all species present
(H3O+ , F-, HF, Cl-, and OH-) in a solution that contains
0.10 M HF 1Ka = 3.5 * 10-42 and 0.10 M HCl.
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Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 149
Sulfur dioxide is quite soluble in water: SO2(g) + H2O(l) ⇌ H2SO3(aq), K = 1.33. The H2SO3 produced is a weak diprotic acid (Ka1 = 1.5 * 10^-2; Ka2 = 6.3 * 10^-8). Calculate the pH and the concentrations of H2SO3, HSO3-, and SO3^2- in a solution prepared by continuously bubbling SO2 at a pressure of 1.00 atm into pure water.
Verified step by step guidance1
Step 1: Use Henry's Law to determine the concentration of dissolved SO2 in water. Henry's Law states that the concentration of a gas in a liquid is proportional to the partial pressure of the gas above the liquid. The formula is C = kH * P, where C is the concentration, kH is the Henry's Law constant for SO2, and P is the pressure of SO2.
Step 2: Recognize that the dissolved SO2 reacts with water to form H2SO3 according to the equilibrium: SO2(g) + H2O(l) ⇌ H2SO3(aq). Use the equilibrium constant K = 1.33 to find the concentration of H2SO3 at equilibrium.
Step 3: Consider the dissociation of H2SO3 in water. Since H2SO3 is a diprotic acid, it dissociates in two steps: H2SO3 ⇌ H+ + HSO3- (with Ka1) and HSO3- ⇌ H+ + SO3^2- (with Ka2). Use the given Ka1 and Ka2 values to set up equilibrium expressions for each dissociation step.
Step 4: Calculate the concentration of H+ ions from the first dissociation step using the expression [H+] = sqrt(Ka1 * [H2SO3]). This will help in determining the pH of the solution.
Step 5: Use the concentrations from the first dissociation to find the concentration of HSO3- and then use the second dissociation equilibrium to find the concentration of SO3^2-. Use the expressions [HSO3-] = [H+] and [SO3^2-] = Ka2 * [HSO3-] / [H+].
Related Practice
Textbook Question
When NO2 is bubbled into water, it is completely converted
to HNO3 and HNO2:
2 NO21g2 + H2O1l2S HNO31aq2 + HNO21aq2
Calculate the pH and the concentrations of all species present
(H3O+ , OH-, HNO2, NO2
-, and NO3
-) in a solution
prepared by dissolving 0.0500 mol of NO2 in 1.00 L of
water. Ka for HNO2 is 4.5 * 10-4.
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Open Question
Normal rain has a pH of 5.6 due to dissolved atmospheric carbon dioxide at a current level of 400 ppm. Various models predict that burning fossil fuels will increase the atmospheric CO2 concentration to between 500 and 1000 ppm by the year 2100. (a) Calculate the pH of rain in a scenario where the CO2 concentration is 750 ppm. CO2 reacts with water to produce carbonic acid according to the equation: CO2(aq) + H2O(l) ⇌ H2CO3(aq). Assume all the dissolved CO2 is converted to H2CO3. Acid dissociation constants for H2CO3 are Ka1 = 4.3 * 10^-7; Ka2 = 5.6 * 10^-11. (Worked Example 16.11 is a model for this calculation.) (b) Will rising CO2 levels affect the acidity of rainfall?
Textbook Question
Acid and base behavior can be observed in solvents other
than water. One commonly used solvent is dimethyl
sulfoxide (DMSO), which can be treated as a monoprotic
acid 'HSol.' Just as water can behave either as an acid or a
base, so HSol can behave either as a Brønsted–Lowry acid
or base.
(b) The weak acid HCN has an acid dissociation constant
Ka = 1.3 * 10-13 in the solvent HSol. If 0.010 mol of
NaCN is dissolved in 1.00 L of HSol, what is the equilibrium
concentration of H2Sol + ?
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Textbook Question
A 7.0 mass % solution of H3PO4 in water has a density of
1.0353 g/mL. Calculate the pH and the molar concentrations
of all species present (H3PO4, H2PO4-, PO43-, H3O+ ,
and OH-) in the solution. Values of equilibrium constants
are listed in Appendix C.
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1
rank
Textbook Question
In the case of very weak acids, 3H3O+ 4 from the dissociation
of water is significant compared with 3H3O+ 4
from the dissociation of the weak acid. The sugar substitute
saccharin 1C7H5NO3S2, for example, is a very weak
acid having Ka = 2.1 * 10-12 and a solubility in water of
348 mg/100 mL. Calculate 3H3O+ 4 in a saturated solution
of saccharin. (Hint: Equilibrium equations for the dissociation
of saccharin and water must be solved simultaneously.)
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