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Chapter 15: Problem 39

Calculate the pH of solutions having the following \(\left[\mathrm{H}^{+}\right]\): (a) \(0.35 \mathrm{M}\) (b) \(1.75 \mathrm{M}\) (c) \(2.0 \times 10^{-5} \mathrm{M}\)

Short Answer

Expert verified
(a) pH = 0.46; (b) pH = -0.24; (c) pH = 4.70

Step by step solution

01

- Understanding the pH Formula

The pH of a solution is calculated using the formula: \[ \text{pH} = -\text{log} \big( [\text{H}^+] \big) \] where \text[ \text{H}^+ \text] is the concentration of hydrogen ions in moles per liter (M).
02

- Calculating pH for 0.35 M

Substitute \text[ \text[ \text[0.35 \text[ in the pH formula: \[ \text{pH} = -\text{log} \big( 0.35 \big) \] Using a calculator, we get: \[ \text{pH} \approx 0.46 \].
03

- Calculating pH for 1.75 M

Substitute \text[ \text[ \text[1.75 \text[ in the pH formula: \[ \text{pH} = -\text{log} \big( 1.75 \big) \] Using a calculator, we get: \[ \text{pH} \approx -0.24 \].
04

- Calculating pH for 2.0 \times 10^{-5} \text[ \text[ M

Substitute \text[ \text[ \text[2.0 \times 10^{-5} \text[ in the pH formula: \[ \text{pH} = -\text{log} \big( 2.0 \times 10^{-5} \big) \] Using a calculator, we get: \[ \text{pH} \approx 4.70 \].

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Key Concepts

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

Hydrogen Ion Concentration
To understand pH calculation, we must first grasp the concept of hydrogen ion concentration \([H^{+}]\). Hydrogen ions heavily influence a solution's acidity or basicity. The more hydrogen ions present, the more acidic the solution. Conversely, fewer hydrogen ions suggest a more basic solution. Concentration is given in moles per liter (M) and is typically expressed in exponential notation, especially for very low concentrations like \(2.0 \times 10^{-5} \text{ M}\). To easily compare different solutions, chemists use the pH scale, which we'll discuss further.
Logarithmic Scale
The pH scale is a logarithmic scale. This means it measures the concentration of hydrogen ions based on their powers of ten. A logarithmic scale condenses large or small numbers into a more manageable range. For example, the difference between \(10^{-5} \text{ M}\) and \(10^{-6} \text{ M}\) is much easier to compare using logarithms.
To calculate pH, we use the formula:\[ \text{pH} = -\text{log} \big( [\text{H}^+] \big) \] The negative sign in the formula indicates that pH decreases as hydrogen ion concentration increases. Understanding logarithms and how they scale numbers is crucial for working with pH values effectively.
Acid-Base Chemistry
Acid-base chemistry is all about the interaction between acids, which donate hydrogen ions, and bases, which accept them.
An acidic solution has a high concentration of hydrogen ions and a low pH, whereas a basic solution has fewer hydrogen ions and a higher pH. The pH values typically range from 0 to 14:
  • pH 0-7: Acidic
  • pH 7: Neutral
  • pH 7-14: Basic
Knowing how to calculate and interpret pH helps in understanding chemical behaviors in solutions, predicting reactions, and ensuring safety in handling chemicals.

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Most popular questions from this chapter

Write an equation to illustrate the acid-base reaction that will take place between perchloric acid, \(\mathrm{HClO}_{4}\), and a hydroxide ion, \(\mathrm{OH}^{-}\). Identify the acids, bases, and their conjugates.

Determine whether each of the following substances is an electrolyte or a nonelectrolyte. All are mixed with water. (a) \(\mathrm{SO}_{3}\) (d) \(\mathrm{O}_{2}\) (insoluble) (b) \(\mathrm{K}_{2} \mathrm{CO}_{3}\) (e) \(\mathrm{CuBr}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{OH}\) (methyl alcohol) (f) \(\mathrm{HI}\)

Determine how many grams of \(\mathrm{Al}(\mathrm{OH})_{3}\) will be required to neutralize \(275 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{HCl}\) according to the reaction: $$ 3 \mathrm{HCl}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$

Calculate the \(\left[\mathrm{H}^{+}\right]\)for (a) pure water (b) lemon lime soda with a \(\mathrm{pH}\) of \(4.3\) (c) root beer with a \(\mathrm{pH}\) of \(5.8\) (d) window cleaner with a \(\mathrm{pH}\) of \(10.3\)

The \(\mathrm{pH}\) of a solution of a strong acid was determined to be 3 . If water is then added to dilute this solution, would the \(\mathrm{pH}\) change? Why or why not? Could enough water ever be added to raise the \(\mathrm{pH}\) of an acid solution above 7 ?
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