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Chapter 5: Problem 51

If you dilute \(25.0 \mathrm{mL}\) of \(1.50 \mathrm{M}\) hydrochloric acid to 500 . mL, what is the molar concentration of the dilute acid?

Short Answer

Expert verified
The molar concentration of the dilute acid is 0.075 M.

Step by step solution

01

Identify the Relationship

To find the molar concentration of the dilute solution, we use the dilution formula, which is \[ C_1V_1 = C_2V_2 \]where \(C_1\) and \(V_1\) are the initial concentration and volume of the solution, and \(C_2\) and \(V_2\) are the final concentration and volume after dilution.
02

Substitute Known Values

We know from the problem statement that \(C_1 = 1.50 \, \text{M}\), \(V_1 = 25.0 \, \text{mL}\), and \(V_2 = 500.0 \, \text{mL}\). Substitute these values into the dilution equation:\[ 1.50 imes 25.0 = C_2 imes 500.0 \]
03

Solve for \(C_2\)

To find \(C_2\), solve the equation \[ 37.5 = C_2 imes 500.0 \]by dividing both sides by 500.0:\[ C_2 = \frac{37.5}{500.0} \]
04

Calculate the New Concentration

Perform the division to find \(C_2\):\[ C_2 = 0.075 \, \text{M} \]This is the molar concentration of the dilute hydrochloric acid.

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Key Concepts

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

Dilution Formula
When dealing with solutions, the dilution formula is a critical tool that helps in calculating changes in concentration. The formula, expressed as \( C_1V_1 = C_2V_2 \), succinctly represents the relationship between concentrations and volumes before and after the dilution process.
This formula works because, in a dilution, the amount of solute remains constant even though the total volume changes.
  • \( C_1 \) and \( V_1 \) refer to the initial concentration and volume of the solution, respectively.
  • \( C_2 \) and \( V_2 \) stand for the new concentration and volume after dilution.
Using this formula, we can calculate how diluting a concentrated solution affects its molarity, which is the concentration of the solution in moles per liter.
Molar Concentration
Molar concentration, or molarity, is an essential aspect of chemistry that defines the number of moles of solute per liter of solution. It is typically expressed in moles per liter (M).
Understanding molarity helps in predicting how a solution will behave in chemical reactions.
**Calculating Molarity** involves the formula:
  • \( \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)
When diluting a solution, the molarity changes as shown in the dilution example, but the total amount of solute stays the same.
Therefore, it is crucial to comprehend how to calculate and adjust molarity in various chemical processes.
Hydrochloric Acid
Hydrochloric acid (HCl) is a highly important mineral acid commonly used in laboratory and industrial settings.
It is a colorless solution and has a pungent smell.
**Properties of Hydrochloric Acid** include:
  • **Corrosive Nature:** It can cause damage to living tissues and materials, which makes handling it with caution important.
  • **Common Uses:** Used in cleaning, digesting foods in stomach, and chemical industry.
  • **Reactivity:** Highly reactive, used in processes like pickling of steel and production of organic compounds.
In the context of preparing dilute solutions from concentrated HCl, understanding the potency and handling precautions of hydrochloric acid is crucial, as it can significantly influence the outcome and safety of experiments and processes.

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Most popular questions from this chapter

You have a bottle of solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and a \(500.0-\mathrm{mL}\). volumetric flask. Explain how you would make a \(0.20 \mathrm{M}\) solution of sodium carbonate.

On General ChemistryNow CD-ROM or website Screen 4.12, Chemical Puzzler, you can explore the reaction of baking soda (NaHCO \(_{3}\) ) with the acetic acid in vinegar. Suppose you place exactly 200 mL. of vinegar in the beaker and add baking soda. The reaction occurring is \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{NaHCO}_{3}(\mathrm{aq}) \longrightarrow\) $$\mathrm{NaCH}_{3} \mathrm{CO}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ How many spoonfuls of baking soda is required to consume the acetic acid in the \(200-\mathrm{mL}\). sample? (Assume there is \(50.0 \mathrm{g}\) of acetic acid per liter of vinegar and a spoonful of baking soda has a mass of \(3.8 \mathrm{g} .\) ) Are three spoonfuls sufficient? Are four spoonfuls enough?

Which two of the following reactions are oxidationreduction reactions? Explain your answer briefly. Classify the remaining reaction. (a) \(\operatorname{Cd} \mathrm{Cl}_{2}(\mathrm{aq})+\mathrm{Na}_{2} \mathrm{S}(\mathrm{aq}) \longrightarrow \operatorname{CdS}(\mathrm{s})+2 \mathrm{NaCl}(\mathrm{aq})\) (b) \(2 \mathrm{Ca}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CaO}(\mathrm{s})\) (c) \(4 \mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{aq})\)

Some potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right), 2.335 \mathrm{g},\) is dissolved in enough water to make exactly \(500 .\) mL of solution. What is the molar concentration of the potassium dichromate? What are the molar concentrations of the \(\mathbf{K}^{+}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ions?

What feature causes the following reactions to be product-favored? (a) \(\mathrm{CuCl}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{S}(\mathrm{aq}) \longrightarrow \mathrm{CuS}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq})\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+3 \mathrm{KOH}(\mathrm{aq}) \longrightarrow 3 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{K}_{3} \mathrm{PO}_{4}(\mathrm{aq})\)
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