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Chapter 8: Problem 216

\(50 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous \(\mathrm{NaOH}\) is mixed with \(10 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous acetic acid. The conc. \(\left(\mathrm{OH}^{-}\right)\)of resulting solution is (a) \(0.04 \mathrm{M}\) (b) \(0.080 \mathrm{M}\) (c) \(0.0607 \mathrm{M}\) (d) \(0.0667 \mathrm{M}\)

Short Answer

Expert verified
The concentration of OH- after mixing is 0.0667 M (choice d).

Step by step solution

01

Calculate Moles of NaOH and Acetic Acid

First, find the moles of NaOH and acetic acid in the solutions before mixing. Moles of NaOH = Volume (L) \( \times \) Molarity = \( 0.050 \, \text{L} \times 0.1 \, \text{M} = 0.005 \, \text{moles} \)Moles of acetic acid = Volume (L) \( \times \) Molarity = \( 0.010 \, \text{L} \times 0.1 \, \text{M} = 0.001 \, \text{moles} \).
02

Determine Limiting Reactant and Excess OH- Ions

NaOH and acetic acid react in a 1:1 ratio to form water and sodium acetate. Since there are more moles of NaOH (0.005 moles) compared to acetic acid (0.001 moles), acetic acid is the limiting reactant. Therefore, 0.001 moles of NaOH will react with 0.001 moles of acetic acid, leaving an excess of 0.004 moles of NaOH.
03

Calculate Total Volume of Solution

The total volume of the solution after mixing is the sum of the volumes of NaOH and acetic acid solutions. Total Volume = 50 mL + 10 mL = 60 mL = 0.060 L.
04

Calculate Concentration of OH- in Solution

After the reaction, the excess NaOH (0.004 moles) will determine the concentration of OH- ions in the solution. Concentration of \( \text{OH}^- \) is given by the number of moles of excess \( \text{OH}^- \) per total volume in liters: \[ \text{Concentration of } \text{OH}^- = \frac{0.004 \text{ moles}}{0.060 \text{ L}} = 0.0667 \text{ M} \]
05

Identify the Correct Answer Choice

Compare the calculated concentration of \(\text{OH}^-\) ions with the choices given. The correct concentration is 0.0667 M, which corresponds to option (d).

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Key Concepts

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

Limiting Reactant
In chemical reactions, the limiting reactant is the substance that gets completely consumed first, stopping the reaction from continuing because there are no more reactants available to react with. By determining which reactant is limiting, you can predict the maximum amount of product that can be formed. In our solution, we mix sodium hydroxide (NaOH) with acetic acid, which react in a 1:1 ratio. Here, acetic acid is the limiting reactant because we only have 0.001 moles of it, which is less than the available 0.005 moles of NaOH.
  • Identify each reactant's quantity in moles.
  • Compare with the required stoichiometric ratio.
  • The smaller amount, based on the ratio, is the limiting reactant.
Stoichiometry
Stoichiometry refers to the precise calculation of reactants and products in chemical reactions. It's based on the balanced chemical equation and helps determine the relationship between the quantities of reactants and products. In our example, we need to determine how NaOH and acetic acid react to form new compounds. For every mole of acetic acid that reacts, one mole of NaOH is used. This stoichiometric relationship tells us how many moles of each reactant are needed and how many moles of product we can expect.
  • Balance the chemical equation.
  • Use coefficients to decide the mole ratio.
  • Calculate the required amount of each component to find the product's amount.
Solution Concentration
Solution concentration indicates how much solute is present in a given quantity of solvent or solution. The unit commonly used for concentration is molarity (M), which is moles of solute per liter of solution. In this exercise, we calculate the concentration of hydroxide ions (OH^-) remaining after the reaction. First, find the total volume of the combined solutions. Here, 50 mL of NaOH is mixed with 10 mL of acetic acid, giving a total of 60 mL (or 0.060 L). Then divide the moles of excess OH^- ions by the total volume to find the final concentration.
  • Calculate the moles of solute left after the reaction.
  • Determine the total volume of the solution.
  • Use the formula: Concentration = Moles/Volume.
Moles Calculation
Moles calculation involves determining the number of molecules or atoms in a substance, using Avogadro's number as a reference. It's fundamental in chemistry to understand how much of a substance is present in a given reaction. Moles are calculated by multiplying the volume of the solution (in liters) by its molarity. For NaOH, we had 50 mL or 0.050 L, which when multiplied by a molarity of 0.1 M, resulted in 0.005 moles. Acetic acid, with 10 mL or 0.010 L, similarly provided us with 0.001 moles using the same process. Understanding moles allows us to compare quantities and perform further calculations easily.
  • Convert volume from mL to L.
  • Apply moles formula: Volume (L) 15 Molarity.
  • Ensure consistent units for easy comparison.

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Most popular questions from this chapter

The solubility product of \(\mathrm{AgCl}\) is \(1.8 \times 10^{-10}\) at \(298 \mathrm{~K}\). The solubility of \(\mathrm{AgCl}\) in \(0.01 \mathrm{M} \mathrm{HCl}\) solution is (a) \(1.8 \times 10^{-8} \mathrm{M}\) (b) \(1.4 \times 10^{-6} \mathrm{M}\) (c) \(1.8 \times 10^{-6} \mathrm{M}\) (d) \(1.4 \times 10^{-4} \mathrm{M}\)

\(\mathrm{M}(\mathrm{OH}) \mathrm{x}\) has \(\mathrm{Ksp}=4 \times 10^{-12}\) and solubility \(10^{-4} \mathrm{M}, \mathrm{x}\) is (a) 1 (b) 2 (c) 3 (d) 4

If molar concentrations of two weak acids are the same, their relative strengths can be compared by (a) \(\frac{\alpha_{1}}{\alpha_{2}}\) (b) \(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\) (c) \(\sqrt{\mathrm{K}_{1} / \mathrm{K}_{2}}\) (d) \(\frac{\left[\mathrm{H}^{+}\right]_{1}}{\left[\mathrm{H}^{+}\right]_{2}}\)

\(500 \mathrm{ml}\) of \(0.2 \mathrm{M} \mathrm{HCl}\) is mixed with \(500 \mathrm{ml}\) of \(0.2 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH} .25 \mathrm{ml}\) of the mixture is titrated with \(0.1\) M \(\mathrm{NaOH}\) solution. By how many units does the \(\mathrm{pH}\) change from the start to the stage when \(\mathrm{HCl}\) is just completely neutralized. \(\mathrm{K}_{\text {, for acetic acid }}=2.0 \times 10^{-5}\). (a) \(3.7\) (b) \(4.4\) (c) \(2.0\) (d) \(3.0\)

When \(0.1 \mathrm{~mol}\) of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\left(\mathrm{~K}_{\mathrm{b}}=5 \times 10^{-4}\right)\) is mixed with \(0.08 \mathrm{~mol}\) of \(\mathrm{HCl}\) and diluted to \(1 \mathrm{~L}\), the \(\mathrm{H}^{+}\)ion concentration in the solution is (a) \(8 \times 10^{-11} \mathrm{M}\) (b) \(6 \times 10^{-5} \mathrm{M}\) (c) \(1.6 \times 10^{-11} \mathrm{M}\) (d) \(8 \times 10^{-2} \mathrm{M}\)
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