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Chapter 14: Problem 58

A solution is prepared by adding 50.0 \(\mathrm{mL}\) of 0.050\(M \mathrm{HBr}\) to 150.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) HI. Calculate \(\left[\mathrm{H}^{+}\right]\) and the \(\mathrm{pH}\) of this solution. HBr and HI are both considered strong acids.

Short Answer

Expert verified
The concentration of H鈦� ions in the mixed solution is 0.0875 M, and the pH of the solution is 1.06.

Step by step solution

01

Identifying the data from the problem

Let's first list down all given information and formulas we are going to use. - Volume of HBr solution: 50.0 mL - Concentration of HBr solution: 0.050 M - Volume of HI solution: 150.0 mL - Concentration of HI solution: 0.10 M - We need to find: [H鈦篯 and pH Since both HBr and HI are strong acids, they dissociate completely in water: \(HBr \rightarrow H^{+} + Br^{-}\) \(HI \rightarrow H^{+} + I^{-}\)
02

Convert volumes to liters

We need the volumes in liters for our calculations. We can convert the given volumes in milliliters (mL) to liters (L) by the following conversions: 1 L = 1000 mL For 50.0 mL of HBr solution: Volume = 50.0 mL * (1 L / 1000 mL) = 0.050 L For 150.0 mL of HI solution: Volume = 150.0 mL * (1 L / 1000 mL) = 0.150 L
03

Calculate the total moles of H鈦� produced from both acids

Next, we calculate the moles of H鈦� ions produced by both HBr and HI solutions using the given concentrations and volumes using the formula: moles of H鈦� = concentration 脳 volume For HBr: moles of H鈦� from HBr = 0.050 M * 0.050 L = 0.0025 mol For HI: moles of H鈦� from HI = 0.10 M * 0.150 L = 0.015 mol Now, find the total moles of H鈦� produced by adding the moles from both acids: Total moles of H鈦� = moles of H鈦� from HBr + moles of H鈦� from HI = 0.0025 mol + 0.015 mol = 0.0175 mol
04

Calculate the final concentration of H鈦� in the solution

The total volume of the solution is the sum of the volumes of HBr and HI solutions: Total volume = volume of HBr + volume of HI = 0.050 L + 0.150 L = 0.200 L Now, calculate the final concentration of H鈦� ions in the solution using the formula: Concentration of H鈦� = (total moles of H鈦�) / (total volume of the solution) [H鈦篯 = 0.0175 mol / 0.200 L = 0.0875 M
05

Calculate the pH of the solution

Now that we have the concentration of H鈦� ions, we can calculate the pH using the formula: pH = -log[H鈦篯 pH = -log(0.0875) = 1.06 Thus, the concentration of H鈦� ions in the mixed solution is 0.0875 M and the pH of the solution is 1.06.

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Key Concepts

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

Strong Acids
Strong acids, such as hydrochloric acid (HCl), hydrobromic acid (HBr), and hydroiodic acid (HI), are characterized by their complete dissociation in water. This means they fully separate into their ions.
For example:
  • For HBr: \( \text{HBr} \rightarrow \text{H}^+ + \text{Br}^- \)
  • For HI: \( \text{HI} \rightarrow \text{H}^+ + \text{I}^- \)
Since they dissociate completely, the concentration of the hydrogen ions \( [\text{H}^+] \) is equal to the initial concentration of the acid in the solution. This property makes calculations involving strong acids more straightforward as the initial acid concentration directly translates to \( [\text{H}^+] \).
Knowing that an acid is strong helps us predict that the solution will be highly acidic.
pH Calculation
The pH of a solution reflects how acidic or basic it is. It is a logarithmic scale used to express the concentration of hydrogen ions. The formula to calculate pH is:
\[ \text{pH} = -\log[\text{H}^+] \]This means that if we know the concentration of hydrogen ions, we can find the pH by taking the negative logarithm (base 10) of that concentration.
A strong acid will have a low pH value because it contributes a high concentration of hydrogen ions to the solution. Remember, as \( [\text{H}^+] \) increases, pH decreases, making the solution more acidic.
H鈦� Concentration
In acid-base chemistry, \([\text{H}^+]\) represents the hydrogen ion concentration. For strong acids like HBr and HI, determining \([\text{H}^+]\) involves calculating the moles of hydrogen ions that result from complete dissociation.
To find the molar concentration of hydrogen ions in the mixed solution:
  • Add up the moles of \( \text{H}^+ \) from each component.
  • Divide the total moles by the total volume of the solution to get \([\text{H}^+]\).
For our exercise, the final concentration was found using:\[ [\text{H}^+] = \frac{0.0175 \, \text{mol}}{0.200 \, \text{L}} = 0.0875 \, \text{M} \]
This concentration reflects the high acidity of the solution due to the strong acids involved.
Molarity Calculations
Molarity is a way to express the concentration of a solution, defined as moles of solute per liter of solution.
In the context of our exercise:
  • Each solution鈥檚 volume is converted from milliliters to liters.
  • The molarity is then used to determine the moles of \(\text{H}^+\).
The formula used is:\[ \text{moles of } \text{H}^+ = \text{molarity} \times \text{volume} \]
By adding the moles of \(\text{H}^+\) from both solutions and dividing by the total volume, we find the final concentration. This step is crucial in blending solutions or when analyzing a solution鈥檚 acid strength.

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Most popular questions from this chapter

Quinine \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, \(\mathrm{p} K_{\mathrm{b}_{1}}=5.1\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.7\left(\mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}\right) .\) Only 1 g quinine will dissolve in 1900.0 \(\mathrm{mL}\) of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction \(\mathrm{Q}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+}+\mathrm{OH}^{-}\) described by \(\mathrm{p} K_{\mathrm{b}_{1}},\) where \(\mathrm{Q}=\) quinine.

Calculate \(\left[\mathrm{OH}^{-}\right], \mathrm{pOH}\) , and pH for each of the following. a. 0.00040\(M \mathrm{Ca}(\mathrm{OH})_{2}\) b. a solution containing 25 \(\mathrm{g}\) KOH per liter c. a solution containing 150.0 \(\mathrm{g}\) NaOH per liter

What are the major species present in a \(0.150-M \mathrm{NH}_{3}\) solution? Calculate the \(\left[\mathrm{OH}^{-}\right]\) and the pH of this solution.

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2},\) is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\) Calculate the pH of a \(0.10-M\) solution of monochloroacetic acid.

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a \(3.0 \times 10^{-7}-\) M solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\)
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