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If \(4.00 \mathrm{mL}\) of \(0.0250 \mathrm{M} \mathrm{CuSO}_{4}\) is diluted to \(10.0 \mathrm{mL}\) with pure water, what is the molar concentration of copper(II) sulfate in the diluted solution?

Short Answer

Expert verified
The molar concentration of copper(II) sulfate in the diluted solution is 0.0100 M.

Step by step solution

01

Understand Dilution Formula

The dilution formula is given by: \[ C_1V_1 = C_2V_2 \] where \( C_1 \) is the initial concentration, \( V_1 \) is the initial volume, \( C_2 \) is the final concentration, and \( V_2 \) is the final volume.
02

Identify Given Values

We know that \( C_1 = 0.0250 \mathrm{M} \), \( V_1 = 4.00 \mathrm{mL} \), and \( V_2 = 10.0 \mathrm{mL} \). We need to find \( C_2 \).
03

Rearrange Dilution Formula

Rearrange the formula to solve for \( C_2 \): \[ C_2 = \frac{C_1V_1}{V_2} \]
04

Substitute Values

Substitute the known values into the equation: \[ C_2 = \frac{0.0250 \times 4.00}{10.0} \]
05

Calculate Final Concentration

Perform the calculation: \[ C_2 = \frac{0.100}{10.0} = 0.0100 \mathrm{M} \].

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Key Concepts

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

Molar Concentration
Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It tells us how many moles of a substance are present in one liter of solution. The unit used for molarity is molar, represented by "M".

To calculate molarity, you can use the formula: \[M = \frac{n}{V}\]where:
  • \(M\) is the molarity in moles per liter (M)
  • \(n\) is the number of moles of solute
  • \(V\) is the volume of the solution in liters
Molarity provides an easy way to discuss concentrations in chemistry that can be used in various calculations, such as determining the amount of reactant needed in a chemical reaction or analyzing dilution in solution preparation.
Copper(II) Sulfate
Copper(II) sulfate, commonly known by its chemical formula \(\text{CuSO}_4\), is a popular chemical compound in labs and industrial applications. It is known for its distinctive blue color when hydrated, as it commonly exists as a pentahydrate \(\text{CuSO}_4\cdot5\text{H}_2\text{O}\). This is the form that appears most often in chemistry laboratories.

Here are several important aspects of Copper(II) sulfate to understand:
  • It is a highly soluble salt in water, making it ideal for various solutions.
  • Frequently used in agriculture and industries for activities such as pest control and electroplating.
  • In educational settings, it's often used to demonstrate chemical reactions and properties, such as conducting specific precipitation reactions.
These properties make copper(II) sulfate an essential component in experiments dealing with solutions and concentrations.
Solution Dilution
Solution dilution is a process whereby the concentration of a solute in a solution is decreased by adding more solvent. It helps chemists and students create solutions of desired concentrations for experiments and various applications.

The fundamental principle behind dilution is that the number of moles of solute remains constant before and after dilution. Consequently, the dilution formula \(C_1V_1 = C_2V_2\) is used to calculate how much solvent is needed or to find the new concentration.

The formula is interpreted as:
  • \(C_1\) and \(V_1\) are the initial concentration and volume, respectively.
  • \(C_2\) is the final concentration we want to find, and \(V_2\) is the final volume after dilution.
This formula allows for the practical application of dilutions in various fields, such as pharmaceuticals, food industry, and biochemistry. Understanding how to apply this concept is crucial for designing experiments and ensuring precise measurements in solution chemistry.

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Most popular questions from this chapter

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

You wish to determine the weight percent of copper in a copper-containing alloy. After dissolving a \(0.251-\mathrm{g}\) sample of the alloy in acid, an excess of KI is added, and the \(\mathrm{Cu}^{2+}\) and \(1^{-}\) ions undergo the reaction $$ 2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq}) $$ The liberated \(I_{3}^{-}\) is titrated with sodium thiosulfate according to the equation \(\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})\) (a) Designate the oxidizing and reducing agents in the two reactions above. (b) If 26.32 mL of \(0.101 M N a_{2} S_{2} O_{3}\) is required for titration to the equivalence point, what is the weight percent of Cu in the alloy?

Potassium hydrogen phthalate, \(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4},\) is used to standardize solutions of bases. The acidic anion reacts with strong bases according to the following net ionic equation: $$ \mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow{\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}(\ell) $$ If a \(0.902-\mathrm{g}\) sample of potassium hydrogen phthalate is dissolved in water and titrated to the equivalence point with \(26.45 \mathrm{mL}\) of \(\mathrm{NaOH},\) what is the molar concentration of the NaOH?

A saturated solution of milk of magnesia, \(\mathrm{Mg}(\mathrm{OH})_{2},\) has a pH of \(10.5 .\) What is the hydronium ion concentration of the solution? Is the solution acidic or basic?

In some laboratory analyses, the preferred technique is to dissolve a sample in an excess of acid or base and then "back-titrate" the excess with a standard base or acid. This technique is used to assess the purity of a sample of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\). Suppose you dissolve a 0.475 -g sample of impure \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in aqueous \(\mathrm{KOH}\) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{KOH}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ The NH ated in the reaction is distilled from the solution into a flask containing \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCl. The ammonia reacts with the acid to produce \(\mathrm{NH}_{4} \mathrm{Cl},\) but not all of the \(\mathrm{HCl}\) is used in this reaction. The amount of excess acid is determined by titrating the solution with standardized NaOH. This titration consumes \(11.1 \mathrm{mL}\) of \(0.121 \mathrm{M} \mathrm{NaOH}\). What is the weight percent of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in the \(0.475-\mathrm{g}\) sample?

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