Chapter 16: Problem 19
\(50.00 \mathrm{mL}\) of \(0.0155 \mathrm{M} \mathrm{HI}(\mathrm{aq})\) is mixed with \(75.00 \mathrm{mL}\) of 0.0106 M KOH(aq). What is the pH of the final solution?
Short Answer
Expert verified
The pH of the final solution is 10.2
Step by step solution
01
Calculate the number of moles for HI and KOH
Using the formula Molarity (M)= moles of solute/liters of solution, calculate moles of HI and KOH. Moles of HI = Molarity * Volume = 0.0155 M * 0.050 L = \(7.75 \times 10^{-4} \, mol\). Moles of KOH = 0.0106 M * 0.075 L = \(7.95 \times 10^{-4} \, mol\)
02
Determine the limiting reactant
In this case, HI is the limiting reactant and KOH is the excess reactant. This is because the moles of HI is less than the moles of KOH.
03
Determine the number of remaining moles of KOH
Calculate the moles of KOH left after reaction with HI. This is obtained by subtracting the moles of HI from the moles of KOH: \(7.95 \times 10^{-4} mol \, - \, 7.75 \times 10^{-4} mol = 2 \times 10^{-5} mol\, KOH\)
04
Calculate the OH- ion concentration in the solution
Since KOH is a strong base, it dissociates completely into K+ and OH- in solution. Therefore, the concentration of OH- ions is equal to the concentration of KOH. The volume of the solution is the sum of the volumes of HI and KOH, which is 0.050 L + 0.075 L = 0.125 L. The concentration \(OH^- = \frac { moles of OH^- }{Volume of Solution } = \frac {2 \times 10^{-5} mol} { 0.125 L} = 1.6 \times 10^{-4} M).
05
Calculate the pOH and then the pH of the solution
The pOH is calculated as \( -\log[OH^-] \, = \, -\log(1.6 \times 10^{-4}) \, = \, 3.8 \). The pH is then calculated as \(14 - pOH \, = \, 14 - 3.8 = 10.2\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molarity
Understanding molarity is crucial when dealing with concentrations in chemistry. Molarity, often symbolized as "M," refers to the concentration of a solution. Specifically, it defines how many moles of a solute are present in one liter of solution.
This concept is fundamental when calculating the amounts of substances involved in reactions. To calculate molarity, you use the formula:
This concept is fundamental when calculating the amounts of substances involved in reactions. To calculate molarity, you use the formula:
- Molarity (M) = moles of solute / liters of solution
Limiting Reactant
The limiting reactant in a chemical reaction is the substance that is completely consumed first, thus determining the extent of the reaction and limiting the amount of product formed.
In the exercise, when mixing HI and KOH, the reaction cannot continue once the HI is used up.
In the exercise, when mixing HI and KOH, the reaction cannot continue once the HI is used up.
- To identify the limiting reactant: calculate the moles of each reactant.
- Compare these to determine which one runs out first.
- Moles of HI = 0.0155 x 0.050 L = 0.000775 mol
- Moles of KOH = 0.0106 x 0.075 L = 0.000795 mol
Acid-Base Reaction
An acid-base reaction is a chemical reaction that occurs between an acid and a base. It typically results in the formation of water and a new compound, often referred to as a salt.
Understanding these types of reactions is essential, particularly in solution chemistry.
In the practice exercise, HI acts as an acid, and KOH serves as a base. Here's what happens:
Understanding these types of reactions is essential, particularly in solution chemistry.
In the practice exercise, HI acts as an acid, and KOH serves as a base. Here's what happens:
- HI donates a proton (H+) to OH- from KOH, generating water (H2O).
- The remaining ions (K+ and I-) form a neutral salt.
pOH and pH Relationship
pH and pOH are concepts that describe the acidity or basicity of a solution. They are inversely related on the pH scale, which ranges from 0 (strongly acidic) to 14 (strongly basic).
Knowing the relationship between pH and pOH is vital for calculating the concentration of hydrogen or hydroxide ions in a solution. Here’s how they relate:
Knowing the relationship between pH and pOH is vital for calculating the concentration of hydrogen or hydroxide ions in a solution. Here’s how they relate:
- The pH is calculated using the formula: \( \text{pH} = -\log[\text{H}^+] \).
- The pOH is calculated as: \( \text{pOH} = -\log[\text{OH}^-] \).
- \( \text{pH} + \text{pOH} = 14 \)