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A \(25.0 \mathrm{mL}\) sample of \(\mathrm{HCl}(\mathrm{aq})\) is diluted to a volume of 500.0 mL. If the concentration of the diluted solution is found to be \(0.085 \mathrm{M} \mathrm{HCl}\), what was the concentration of the original solution?

Short Answer

Expert verified
The concentration of the original solution was \(1.7 \mathrm{M} \mathrm{HCl}\).

Step by step solution

01

Understanding the Given Information

The initial volume of \( \mathrm{HCl} \) solution is \( 25.0 \mathrm{mL} \) or \( 0.0250 \mathrm{L} \). The diluted volume is \( 500.0 \mathrm{L} \). The molarity of the diluted solution (M2) is \( 0.085 \mathrm{M} \). We need to find the molarity of the initial solution (M1).
02

Use the Dilution Formula

The dilution formula is \( M1V1 = M2V2 \). Inserting the given values into this equation gives \( M1(0.0250L) = 0.085M(0.500L) \).
03

Solve for M1

We solve the equation for M1 by dividing both sides of the equation by \( 0.0250L \). That gives \( M1 = \frac{0.085M(0.500L)}{0.0250L} \).
04

Calculate the Result

Performing the calculation gives \( M1 = 1.7M \). So, the molarity of the original solution was \( 1.7M \) .

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Key Concepts

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

HCl solution concentration
To understand how concentrated a hydrochloric acid (HCl) solution is, we need to discuss what concentration means. Concentration is a way of expressing how much of a substance (solute) is present in a certain volume of solution. For HCl solutions, concentration is typically indicated in molarity (M), which is moles of HCl per liter of solution.
For example, if you have 1 mole of HCl dissolved in 1 liter of solution, this would be a 1 M HCl solution. Understanding the concentration of any solution is crucial because it affects how the substance will react in experiments or chemical processes. Steps like dilution, where the concentration is adjusted, are commonly performed in labs to achieve the desired strength of a solution.
When you dilute a solution like HCl, you're adding more solvent, typically water, which decreases the concentration of the solute throughout the new total volume of the solution.
dilution formula
The dilution formula is a quick way to figure out how the concentration of a solution changes when you add more solvent. It's expressed as \(M1V1 = M2V2\). Here:
  • \(M1\) is the molarity (concentration) of the original solution.
  • \(V1\) is the volume of the original solution.
  • \(M2\) is the molarity of the diluted solution.
  • \(V2\) is the total volume after dilution.
This formula arises from the principle of conservation of moles. The number of moles before dilution (\(M1V1\)) is equal to the number of moles after dilution (\(M2V2\)).
The key idea is that while adding solvent changes the volume and concentration, it does not change the total number of moles of the solute in solution.
molarity calculation
Molarity is one of the most common ways to express solution concentration in chemistry. Calculating molarity involves determining how many moles of solute are dissolved in one liter of solution. The formula for molarity is \(M = \frac{n}{V}\), where:
  • \(M\) is the molarity.
  • \(n\) is the number of moles of the solute.
  • \(V\) is the volume of the solution in liters.
To calculate molarity, you'll need to know some additional information, like the mass of the solute and its molar mass to find \(n\). Or, as in dilution problems, you might know the final volume and concentration and back-calculate using the dilution formula.
Simplifying molarity calculations often requires you to convert measurements correctly—for instance, converting milliliters to liters by dividing by 1000. This ensures consistency and accuracy in calculations, crucial for experiments and applications in chemistry.

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Most popular questions from this chapter

Excess \(\mathrm{NaHCO}_{3}\) is added to \(525 \mathrm{mL}\) of \(0.220 \mathrm{M}\) \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} .\) These substances react as follows: \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NaHCO}_{3}(\mathrm{s}) \longrightarrow\) $$ \mathrm{CuCO}_{3}(\mathrm{s})+2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) How many grams of the \(\mathrm{NaHCO}_{3}(\mathrm{s})\) will be consumed? (b) How many grams of \(\mathrm{CuCO}_{3}(\mathrm{s})\) will be produced?

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