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Chapter 2: Problem 5

Calculation of Hydrogen Ion Concentration from \(\mathrm{pH}\) What is the \(\mathrm{H}^{+}\)concentration of a solution with \(\mathrm{pH}\) of a. \(3.82\); b. \(6.52\); c. \(11.11\) ?

Short Answer

Expert verified
a. \(1.51 \times 10^{-4}\, \text{mol/L}\); b. \(3.02 \times 10^{-7}\, \text{mol/L}\); c. \(7.76 \times 10^{-12}\, \text{mol/L}\).

Step by step solution

01

Understanding the Concept of pH

The pH of a solution is a measure of its acidity or alkalinity, which is defined as the negative logarithm of the hydrogen ion concentration. Mathematically, it is expressed as: \[pH = -\log[\text{H}^{+}] \]where \([\text{H}^{+}]\) represents the concentration of hydrogen ions in moles per liter (mol/L). This relationship will help us find the \([\text{H}^{+}]\) from a given pH.
02

Rearranging the pH Formula

Using the equation: \[pH = -\log[\text{H}^{+}] \]we need to solve for \([\text{H}^{+}]\). To do this, we rearrange the formula to isolate \([\text{H}^{+}]\). This involves taking the antilog (or inverse logarithm) of both sides of the equation:\[[\text{H}^{+}] = 10^{-\text{pH}}\]This equation allows us to calculate the hydrogen ion concentration from a given pH.
03

Calculating [H+] for pH = 3.82

Substitute \(\text{pH} = 3.82\) into the formula \([\text{H}^{+}] = 10^{-\text{pH}}\):\[[\text{H}^{+}] = 10^{-3.82}\]Using a calculator, we find:\[[\text{H}^{+}] \approx 1.51 \times 10^{-4} \text{ mol/L}\]This is the hydrogen ion concentration for a solution with a pH of 3.82.
04

Calculating [H+] for pH = 6.52

Substitute \(\text{pH} = 6.52\) into the formula \([\text{H}^{+}] = 10^{-\text{pH}}\):\[[\text{H}^{+}] = 10^{-6.52}\]Using a calculator, we find:\[[\text{H}^{+}] \approx 3.02 \times 10^{-7} \text{ mol/L}\]This is the hydrogen ion concentration for a solution with a pH of 6.52.
05

Calculating [H+] for pH = 11.11

Substitute \(\text{pH} = 11.11\) into the formula \([\text{H}^{+}] = 10^{-\text{pH}}\):\[[\text{H}^{+}] = 10^{-11.11}\]Using a calculator, we find:\[[\text{H}^{+}] \approx 7.76 \times 10^{-12} \text{ mol/L}\]This is the hydrogen ion concentration for a solution with a pH of 11.11.

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Key Concepts

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

Understanding Hydrogen Ion Concentration
The term "hydrogen ion concentration" refers to the amount of hydrogen ions (H鈦�) present in a solution. These ions are a key player in determining the solution's acidity. The concentration of hydrogen ions is expressed in moles per liter (mol/L).
- A high concentration of hydrogen ions usually indicates an acidic solution, whereas a low concentration signifies a basic or alkaline solution.
To compute hydrogen ion concentration from pH, we rely on the formula:\[[H^{+}] = 10^{-pH}\]This formula allows us to take the negative exponent of the pH value to determine the exact concentration of hydrogen ions. This method is essential when assessing the nature of a substance, especially in chemical reactions and environmental science.
The Role of Inverse Logarithm in pH Calculation
Inverse logarithm might sound complicated, but it simply reverses the logarithmic function. In the context of pH calculation, pH is originally defined by the equation:\[pH = -\log{[H^{+}]}\]To find the hydrogen ion concentration from a given pH, we apply the inverse of this logarithm. This process involves rearranging the pH equation to solve for H鈦�. Instead of thinking in terms of base-ten logarithms directly, we can consider the inversion as the reverse鈥攖he antilog. The equation is then transformed as:\[[H^{+}] = 10^{-pH}\]The inverse logarithm thus helps convert a pH value into a tangible measure of hydrogen concentration. Understanding this reversal is crucial for quick and accurate chemical analyses.
Balancing Acidity and Alkalinity
Acidity and alkalinity describe the overall character of a solution based on its hydrogen ion concentration. The pH scale鈥攔anging from 0 to 14鈥攊s our best friend for determining this balance:
- A pH less than 7 signals an acidic solution. - A pH of 7 means the solution is neutral, like pure water. - A pH greater than 7 indicates an alkaline (basic) solution.
Each unit change in pH represents a tenfold change in hydrogen ion concentration. This factor highlights the logarithmic nature of the scale, where small numerical changes equate to significant shifts in acidity or alkalinity.
Understanding acidity and alkalinity is vital in diverse fields, from biology to industrial applications, as it dictates the behavior of substances and their interactions in various environments.

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Most popular questions from this chapter

Working with Buffers A buffer contains \(0.010\) mol of lactic acid \(\left(\mathrm{p} K_{\mathrm{n}}=3.56\right)\) and \(0.050\) mol of sodium lactate per liter. a. Calculate the pH of the buffer. b. Calculate the change in \(\mathrm{pIl}\) after adding \(5.0 \mathrm{~mL}\) of \(0.5\) M MCl to 1 L of the huffer. c. What pH change would you expect if you added the same quantity of HCl to 1 L of pure water?

Biological Advantage of Weak Interactions The associations between biomolecules are often stabilized by hydrogen bonds, electrostatic interactions, the hydrophobic effect, and van der Waals interactions. How are weak interactions such as these advantageous to an organism?

Identifying Conjugate Bases Write the conjugate base for each acid: a. \(\mathrm{H}_{3} \mathrm{PO}_{4}\) b. \(\mathrm{H}_{2} \mathrm{CO}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{COOH}\) d. \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\)

Calculation of Blood pH from \(\mathrm{CO}_{2}\) and Baicarbonate Levels Calculate the \(\mathrm{pH}\) of a blood plasma sample with a total \(\mathrm{CO}_{2}\) concentration of \(26.9 \mathrm{~mm}\) and bicarbonate concentration of \(25.6 \mathrm{~mm}\). Recall from page 63 that the relevant \(\mathrm{p} K_{\mathrm{a}}\) of carbonic acid is \(6.1\).

Calculation of the \(\mathrm{pH}\) of a Mixture of a Weak Acid and Its Conjugate Base Calculate the \(\mathrm{pH}\) of a dilute solution that contains a molar ratio of potassium acetate to acetic acid \(\left(\mathrm{p} K_{\mathrm{a}}=4.76\right)\) of a. \(2: 1\) b. \(1: 3 ;\) c. \(5: 1\) d. \(1: 1 ;\) e. \(1: 10\).
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