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Chapter 14: Problem 57

Calculate the \(\mathrm{pH}\) of each of the following solutions of a strong acid in water. a. \(0.10 \mathrm{MHCl}\) c. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\) b. \(5.0 \mathrm{M} \mathrm{HCl}\)

Short Answer

Expert verified
The pH of the given solutions are: a) For the \(0.10 M\) HCl solution, the pH is 1. c) For the \(1.0 \times 10^{-11} M\) HCl solution, the pH is 11. b) For the \(5.0 M\) HCl solution, the pH is approximately -0.3.

Step by step solution

01

Determine the concentration of H+ ions

Since HCl is a strong acid, it will dissociate completely in water. So, the concentration of H+ ions will be equal to the concentration of HCl in the solution. For this solution, [H+] = \([0.10 M]\).
02

Calculate the pH of the solution

To calculate the pH, use the formula: pH = -log[H+]. So, for this solution, pH = - log\[0.10\] = 1. For solution c:
03

Determine the concentration of H+ ions

Since HCl is a strong acid, it will dissociate completely in water. So, the concentration of H+ ions will be equal to the concentration of HCl in the solution. For this solution, [H+] = \(1.0 \times 10^{-11} M\).
04

Calculate the pH of the solution

To calculate the pH, use the formula: pH = -log[H+]. So, for this solution, pH = - log\[\(1.0 \times 10^{-11}\)\] = 11. For solution b:
05

Determine the concentration of H+ ions

Since HCl is a strong acid, it will dissociate completely in water. So, the concentration of H+ ions will be equal to the concentration of HCl in the solution. For this solution, [H+] = \([5.0 M]\).
06

Calculate the pH of the solution

To calculate the pH, use the formula: pH = -log[H+]. So, for this solution, pH = - log\[5.0\]. Since 5.0 is between 1 and 10, the pH will be between 0 and 1. We can further estimate it as pH 鈮� -0.3.

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Key Concepts

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

Strong Acids
Strong acids are unique because they completely dissociate in water. This means they break apart entirely into their respective ions.
For instance, when hydrochloric acid (HCl) is added to water, it separates entirely into hydrogen ions (H鈦�) and chloride ions (Cl鈦�).
This complete dissociation is what defines strong acids.
  • They have no remaining intact acid molecules in solution.
  • Examples besides HCl include nitric acid (HNO鈧�) and sulfuric acid (H鈧係O鈧�).
Because of this property, the concentration of the acid directly equals the concentration of hydrogen ions.
Understanding strong acids is crucial for pH calculations because it simplifies the process.
HCl Dissociation
Hydrochloric acid, or HCl, is a common example used to explain the dissociation of strong acids.
In water, HCl completely dissociates into hydrogen ions (H鈦�) and chloride ions (Cl鈦�). This process can be represented by the equation:
\[\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]When calculating pH, we can assume that the molarity of H鈦� ions in solution is the same as the molarity of the original HCl solution at the start.
This straightforward conversion is possible because of the full dissociation.
  • It's a single-step dissociation as it forms only two ions.
  • This complete break means that each HCl molecule produces one H鈦� ion.
This clarity helps simplify pH calculations, especially when considering solutions with different concentrations of HCl.
Logarithmic Scale
The pH scale is logarithmic, which means each whole pH value below 7 is ten times more acidic than the next higher value.
For example, a pH of 3 is ten times more acidic than a pH of 4.
This logarithmic nature means small changes in pH indicate significant changes in acidity.
  • pH is calculated using the formula: \( \text{pH} = -\log [\text{H}^+] \)
  • The logarithm comes from the powers of ten in hydrogen ion concentration.
Understanding this concept is critical when moving between different strengths of acid solutions and observing how sensitive the pH scale is to changes in hydrogen ion concentration.
Concentration of Hydrogen Ions
The concentration of hydrogen ions \([\text{H}^+]\) in a solution directly affects its pH.
For strong acids like HCl, this concentration is straightforward to determine as it matches the initial concentration of the acid due to full dissociation.
  • Higher \([\text{H}^+]\) results in a lower pH, indicating higher acidity.
  • Conversely, lower \([\text{H}^+]\) leads to a higher pH, suggesting lower acidity.
When calculating pH, especially for a strong acid like HCl, determining \([\text{H}^+]\) is often the first step since it heavily influences the final pH value.
Accurate measurement of this concentration is key in many chemical applications, illustrating the importance of understanding and calculating \([\text{H}^+]\) efficiently.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

An unknown salt is either \(\mathrm{NaCN}, \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), NaF, \(\mathrm{NaCl}\), or \(\mathrm{NaOCl}\). When \(0.100\) mole of the salt is dissolved in \(1.00 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is \(8.07\). What is the identity of the salt?

A \(0.15-M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}}\)

Calculate the \(\mathrm{pH}\) of the following solutions: a. \(1.2 M \mathrm{CaBr}_{2}\) b. \(0.84 \mathrm{MC}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}}\right.\) for \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\) ) c. \(0.57 \mathrm{M} \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}\right.\) for \(\left.\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)\)

A \(10.0\) -mL sample of an \(\mathrm{HCl}\) solution has a \(\mathrm{pH}\) of \(2.000\). What volume of water must be added to change the \(\mathrm{pH}\) to \(4.000\) ?
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