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Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 16.85

Calculate the pH of solutions prepared by:
(d) Mixing equal volumes of 0.20 M HCl and 0.50 M HNO3.
(Assume that volumes are additive.)

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1
Identify the nature of the acids: Both HCl and HNO3 are strong acids, meaning they completely dissociate in water.
Calculate the moles of HCl: Use the concentration (0.20 M) and assume a volume V to find moles of HCl as 0.20V.
Calculate the moles of HNO3: Use the concentration (0.50 M) and the same volume V to find moles of HNO3 as 0.50V.
Determine the total moles of H+ ions: Since both acids dissociate completely, add the moles of H+ from HCl and HNO3, which is 0.20V + 0.50V.
Calculate the concentration of H+ ions in the final solution: Since the volumes are additive, the total volume is 2V. The concentration of H+ is (0.20V + 0.50V) / 2V.
Determine the pH: Use the formula pH = -log[H+] to find the pH of the solution.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

pH Scale

The pH scale measures the acidity or basicity of a solution, ranging from 0 to 14. A pH of 7 is neutral, while values below 7 indicate acidity and above 7 indicate alkalinity. The pH is calculated as the negative logarithm of the hydrogen ion concentration, [H+], in moles per liter. Understanding the pH scale is essential for determining the acidity of mixed solutions.
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Strong Acids

Strong acids, such as HCl and HNO3, completely dissociate in water, releasing all their hydrogen ions into the solution. This complete ionization means that the concentration of hydrogen ions directly corresponds to the concentration of the acid. When mixing solutions of strong acids, the total concentration of hydrogen ions can be calculated by summing the contributions from each acid, which is crucial for determining the resulting pH.
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Strong Acid-Strong Base Titration

Dilution and Volume Addition

When mixing solutions, the total volume of the resulting solution affects the concentration of solutes. Assuming volumes are additive means that the final volume is the sum of the individual volumes. This principle is important for calculating the new concentrations of the acids after mixing, as the dilution effect will influence the final pH of the solution.
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Dilution Equation
Related Practice
Open Question
For each of the Lewis acid–base reactions in Problem 16.138, draw electron-dot structures for the reactants and products, and use the curved arrow notation to represent the donation of a lone pair of electrons from the Lewis base to the Lewis acid.
Open Question
For a solution of two weak acids with comparable values of Ka, there is no single principal reaction. The two acid dissociation equilibrium equations must therefore be solved simultaneously. Calculate the pH in a solution that is 0.10 M in acetic acid (CH3CO2H, Ka = 1.8 * 10^-5) and 0.10 M in benzoic acid (C6H5CO2H, Ka = 6.5 * 10^-5). (Hint: Let x = [CH3CO2H] that dissociates and y = [C6H5CO2H] that dissociates; then [H3O+] = x + y.)
Textbook Question

What is the pH and the principal source of H3O+ ions in 1.0 * 10-10 M HCl? (Hint: The pH of an acid solution can’t exceed 7.) What is the pH of 1.0 * 10-7 M HCl?

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Textbook Question

The acidity of lemon juice is derived primarily from citric acid (H3Cit), a triprotic acid. What are the concentrations of H3Cit, H2Cit-, HCit2-, and Cit3- in a sample of lemon juice that has a pH of 2.37 and a total concentration of the four citrate-containing species of 0.350 M?

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Textbook Question

A 100.0 mL sample of a solution that is 0.100 M in HCl and 0.100 M in HCN is titrated with 0.100 M NaOH. Calculate the pH after the addition of the following volumes of NaOH:

(a) 0.0 mL

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Textbook Question

A 40.0 mL sample of a mixture of HCl and H3PO4 was titrated with 0.100 M NaOH. The first equivalence point was reached after 88.0 mL of base, and the second equivalence point was reached after 126.4 mL of base.

(e) Sketch the pH titration curve, and label the buffer regions and equivalence points.

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