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Chapter 14: Problem 61

Calculate the concentration of an aqueous HBr solution that has \(\mathrm{pH}=4.25 . \mathrm{HBr}\) is a strong acid.

Short Answer

Expert verified
The concentration of the aqueous HBr solution is \(5.623 \times 10^{-5} \mathrm{M}\).

Step by step solution

01

Write the dissociation equation of HBr.

For this task, we must first write the equation for the dissociation of HBr in the aqueous solution. Since HBr is a strong acid, it will dissociate completely as follows: \[ \mathrm{HBr} \rightarrow \mathrm{H}^{+} + \mathrm{Br}^- \]
02

Understand the relationship between pH and the concentration of \(\mathrm{H}^{+}\) ions.

The pH is defined as follows: \[ \mathrm{pH} = -\log_{10} [\mathrm{H}^+] \] Where \(\mathrm{pH}\) is the pH value and \([\mathrm{H}^+]\) is the concentration of \(\mathrm{H}^{+}\) ions in the solution. The \(\mathrm{pH}\) given in our problem is 4.25.
03

Solve for the concentration of \(\mathrm{H}^{+}\) ions.

To find the concentration of \(\mathrm{H}^{+}\) ions, we will use the given pH value and the pH formula: \[ 4.25 = -\log_{10} [\mathrm{H}^+] \] To solve this equation for \([\mathrm{H}^{+}]\) we must perform the following steps: 1. Multiply both sides of the equation by -1 to isolate the logarithm on the right side of the equation: \[ -4.25 = \log_{10} [\mathrm{H}^+] \] 2. Now, we need to rewrite the equation in exponent form to solve for \([\mathrm{H}^{+}]\): \[ [\mathrm{H}^+] = 10^{-4.25} \] 3. Calculate the concentration of \(\mathrm{H}^{+}\) ions: \[ [\mathrm{H}^+] = 5.623 \times 10^{-5} \mathrm{M} \]
04

Determine the concentration of the HBr solution.

Since HBr is a strong acid and dissociates completely into \(\mathrm{H}^{+}\) ions and \(\mathrm{Br}^-\) ions, the concentration of the HBr solution is equal to the concentration of the \(\mathrm{H}^{+}\) ions: \[ [\mathrm{HBr}] = [\mathrm{H}^+] = 5.623 \times 10^{-5} \mathrm{M} \] Hence, the concentration of the aqueous HBr solution is \(5.623 \times 10^{-5} \mathrm{M}\).

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What mass of KOH is necessary to prepare 800.0 \(\mathrm{mL}\) of a solution having a \(\mathrm{pH}=11.56 ?\)

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