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Chapter 15: Problem 42

Calculate the \(p \mathrm{H}\) of \(6.9 \times 10^{-8} \mathrm{MHBr}\)

Short Answer

Expert verified
Answer: The pH of the HBr solution is approximately 7.16.

Step by step solution

01

Identify the strong acid and its dissociation in water.

In this case, hydrobromic acid (HBr) is a strong acid, which means it will dissociate completely in water as follows: \(\mathrm{HBr} \rightarrow \mathrm{H^+} + \mathrm{Br^-}\) Since we are given the concentration of HBr, we can easily determine the concentration of \(\mathrm{H^+}\) ions in the solution.
02

Recall the formula for pH calculation

The formula for pH is given as: pH = \(- \log_{10}[\mathrm{H^+}]\) Where \([\mathrm{H^+}]\) is the concentration of hydrogen ions in moles per liter (\(\mathrm{M}\))
03

Calculate the pH of the solution

We are given the concentration of HBr: \(6.9 \times 10^{-8} \mathrm{M}\) Since HBr is a strong acid, it dissociates completely in water. Thus, the concentration of \(\mathrm{H^+}\) ions will be the same as the concentration of HBr. Now, we can plug in the value of \([\mathrm{H^+}]\) in the pH formula: pH = \(- \log_{10}(6.9 \times 10^{-8})\) Using a calculator, we get: pH = 7.16
04

Interpret the result

The pH of the \(6.9 \times 10^{-8} \mathrm{MHBr}\) solution is approximately 7.16. Since the pH value is greater than 7, the solution is slightly basic.

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Key Concepts

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

Acid Dissociation
When you dissolve hydrobromic acid (HBr) in water, it undergoes a process called dissociation. Dissociation is when molecules split into smaller particles, like ions. For HBr, this means breaking down into hydrogen ions (\(\mathrm{H^+}\)) and bromide ions (\(\mathrm{Br^-}\)). This happens completely because HBr is considered a strong acid which means it breaks down entirely in water.
  • This complete dissociation results in the concentration of hydrogen ions being equal to the initial concentration of the acid in the solution.
  • Knowing this is crucial for calculating the next steps, as it simplifies finding the hydrogen ion concentration directly from the given acid concentration.
Therefore, with a strong acid like HBr, you can confidently assume the number of hydrogen ions is equal to the amount of HBr dissolved initially.
Hydrogen Ion Concentration
Hydrogen ion concentration determines a solution's acidity. For strong acids like HBr, the concentration of hydrogen ions (\([\mathrm{H^+}]\)) is the same as the acid's initial concentration because they dissociate completely. In this exercise, the given concentration of HBr is \(6.9 \times 10^{-8} \mathrm{M}\).
Therefore, the concentration of \(\mathrm{H^+}\) ions in the solution is also \(6.9 \times 10^{-8} \mathrm{M}\).
  • Knowing the concentration of hydrogen ions is vital for calculating the pH of a solution.
  • It helps us understand the degree of acidity or basicity of the solution.
With this concentration in hand, you are ready to find out the solution's pH level.
pH Formula
To determine how acidic or basic a solution is, you need to calculate the pH. The formula is simple:\[\text{pH} = -\log_{10}([\mathrm{H^+]})\]Where \([\mathrm{H^+}]\) is the concentration of hydrogen ions in moles per liter.
  • This formula takes the log of the hydrogen ion concentration and multiplies it by -1 to convert the positive hydrogen ion concentration into the more intuitive pH scale, where lower numbers represent higher acidity.
  • A pH of 7 is neutral, values <7 are acidic, and >7 are basic.
In the given exercise, using \(6.9 \times 10^{-8} \mathrm{M}\) as the concentration of \(\mathrm{H^+}\) ions, we can calculate:\[\text{pH} = -\log_{10}(6.9 \times 10^{-8})\]This gives us a pH value of approximately 7.16, which seems neutral, yet indicating slight basicity due to minimal but important factors like water's ionic product.

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Most popular questions from this chapter

Identify the acids and bases in the following reactions: a. \(\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \rightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)\) b. \(\mathrm{MgCO}_{3}(s)+2 \mathrm{HCl}(a q) \rightarrow\) \(\mathrm{MgCl}_{2}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell)\) c. \(2 \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(a q)\)

Identify the conjugate base of each of the following compounds: \(\mathrm{HNO}_{2}, \mathrm{HClO}, \mathrm{H}_{3} \mathrm{PO}_{4},\) and \(\mathrm{NH}_{3}\)

Compounds that do not ionize in water have been known to ionize in nonaqueous solvents. In such a solvent, what would be the conjugate acid and conjugate base of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) ?

Painkillers Morphine is an effective painkiller but is also highly addictive. Codeine is a popular prescription painkiller because it is much less addictive than morphine. Codeine contains a basic nitrogen atom that can be protonated to give the conjugate acid of codeine. a. Calculate the \(\mathrm{pH}\) of \(1.8 \times 10^{-3} \mathrm{M}\) morphine if its \(\mathrm{p} K_{\mathrm{b}}\) \(=5.79\) b. Calculate the \(\mathrm{pH}\) of \(2.7 \times 10^{-4} M\) codeine if the \(\mathrm{p} K_{\mathrm{a}}\) of the conjugate acid is 8.21

Which of the following salts produces an acidic solution in water: ammonium acetate, ammonium nitrate, or sodium formate?
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