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Chapter 4: Problem 121

Suppose you dilute \(25.0 \mathrm{mL}\) of a \(0.110 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to exactly \(100.0 \mathrm{mL}\). You then take exactly \(10.0 \mathrm{mL}\) of this diluted solution and add it to a 250-mL volumetric flask. After filling the volumetric flask to the mark with distilled water (indicating the volume of the new solution is \(250 . \mathrm{mL}\) ), what is the concentration of the diluted \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) solution?

Short Answer

Expert verified
The final concentration is \(0.00110 \mathrm{M}\).

Step by step solution

01

Calculate Initial Moles of Na2CO3

To find the number of moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) in the initial solution, use the molarity formula: \[\text{moles} = \text{molarity} \times \text{volume (in L)}\] The initial solution is \(25.0 \mathrm{mL}\) or \(0.0250 \mathrm{L}\) with a concentration of \(0.110 \mathrm{M}\). \[\text{moles} = 0.110 \times 0.0250 = 0.00275 \text{ moles}\]
02

Calculate Concentration After Initial Dilution

Dilute the \(0.00275\) moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to \(100.0 \mathrm{mL}\) or \(0.100 \mathrm{L}\). Use the formula: \[\text{molarity} = \frac{\text{moles}}{\text{volume in L}}\] So, \[\text{molarity} = \frac{0.00275}{0.100} = 0.0275 \mathrm{M}\]
03

Calculate Moles in 10.0 mL of Diluted Solution

From the concentrated solution, calculate the moles in \(10.0 \mathrm{mL}\) or \(0.0100 \mathrm{L}\): \[\text{moles} = 0.0275 \times 0.0100 = 0.000275 \text{ moles}\]
04

Calculate Concentration in 250 mL Solution

Now, this \(0.000275\) moles is diluted to a total volume of \(250 \mathrm{mL}\) or \(0.250 \mathrm{L}\). Calculate the new concentration: \[\text{molarity} = \frac{0.000275}{0.250} = 0.00110 \mathrm{M}\]

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Key Concepts

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

Dilution
Dilution involves reducing the concentration of a solute in a solution, often by adding more solvent. Imagine making lemonade less sour by adding water鈥攖his is dilution. The initial number of moles of solute remains constant; only the volume changes.

Use the formula \( C_1 V_1 = C_2 V_2 \), where \( C_1 \) and \( V_1 \) are the initial concentration and volume, and \( C_2 \) and \( V_2 \) are the final concentration and volume, respectively.
  • Volume and concentration are inversely related. Increase in volume leads to a decrease in concentration.
  • The process maintains the same amount of solute, perfect for laboratory experiments requiring specific concentrations.
Understanding dilution helps in accurately preparing chemical solutions with desired strengths.
Molarity Calculations
Molarity measures the concentration of a solution, expressed in moles of solute per liter of solution (\( ext{M} \)). It's calculated using the formula:\[ ext{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \]In molarity calculations, it's essential to convert all volumes to liters and use proper significant figures.

Steps for calculating molarity:
  • Find the number of moles using the initial concentration and volume.
  • Divide the moles by the total volume of the solution after dilution.
This concept is crucial for preparing solutions with precise concentrations for experiments and industrial applications.
Chemical Solutions
Chemical solutions are homogeneous mixtures of two or more substances. In a solution, the solute is the substance being dissolved, like salt in water. The solvent is the substance in which the solute dissolves, typically more abundant.

Characteristics of chemical solutions:
  • Uniform composition throughout.
  • Solute and solvent can't be separated by filtration.
  • Clear and do not scatter light significantly.
Understanding solutions is fundamental in chemistry, affecting everything from reaction rates to product formation in both natural and industrial processes. Solutions facilitate chemical reactions by allowing molecules to interact in a dissolved state, thereby increasing the efficiency and speed of these interactions.

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Most popular questions from this chapter

Sulfuric acid can be prepared starting with the sulfide ore, cuprite \(\left(\mathrm{Cu}_{2} \mathrm{S}\right) .\) If each \(\mathrm{S}\) atom in \(\mathrm{Cu}_{2} \mathrm{S}\) leads to one molecule of \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what is the theoretical yield of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from \(3.00 \mathrm{kg}\) of \(\mathrm{Cu}_{2} \mathrm{S}\) ?

The nitrite ion is involved in the biochemical nitrogen cycle. You can determine the nitrite ion content of a sample using spectrophotometry by first using several organic compounds to form a colored compound from the ion. The following data were collected. $$\begin{array}{|c|c|}\hline \begin{array}{c}\mathrm{NO}_{2}^{-} \text {lon } \\\\\text { Concentration }\end{array} & \begin{array}{c}\text { Absorbance of } \\\\\text { Solution at } 550 \mathrm{nm}\end{array} \\\\\hline 2.00 \times 10^{-6} \mathrm{M} & 0.065 \\\6.00 \times 10^{-6} \mathrm{M} & 0.205 \\\10.00 \times 10^{-6} \mathrm{M} & 0.338 \\ 14.00 \times 10^{-6} \mathrm{M} & 0.474 \\\\\hline 18.00 \times 10^{-6} \mathrm{M} & 0.598 \\\\\hline \text { Unknown solution } & 0.402 \\\\\hline\end{array}$$ (a) Construct a calibration plot, and determine the slope and intercept. (b) What is the nitrite ion concentration in the unknown solution?

A solution of hydrochloric acid has a volume of \(125 \mathrm{mL}\) and a pH of \(2.56 .\) What mass of \(\mathrm{NaHCO}_{3}\) must be added to completely consume the HCl?

Anhydrous calcium chloride is a good drying agent because it will rapidly pick up water. Suppose you have stored some carefully dried \(\mathrm{CaCl}_{2}\) in a desiccator. Unfortunately, someone did not close the top of the desiccator tightly, and the \(\mathrm{CaCl}_{2}\) became partially hydrated. A \(150-\mathrm{g}\) sample of this partially hydrated material was dissolved in \(80 \mathrm{g}\) of hot water. When the solution was cooled to \(20^{\circ} \mathrm{C}, 74.9 \mathrm{g}\) of \(\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) precipitated. Knowing the solubility of calcium chloride in water at \(20^{\circ} \mathrm{C}\) is \(74.5 \mathrm{g} \mathrm{CaCl}_{2} / 100 \mathrm{g}\) water, determine the water content of the 150 -g sample of partially hydrated calcium chloride (in moles of water per mole of \(\mathrm{CaCl}_{2}\) ).

Boron forms a series of compounds with hydrogen, all with the general formula \(\mathrm{B}_{x} \mathrm{H}_{y}\). $$\mathrm{B}_{x} \mathrm{H}_{y}(\mathrm{s})+\text { excess } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \frac{x}{2} \mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{s})+\frac{\gamma}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ If \(0.148 \mathrm{g}\) of one of these compounds gives \(0.422 \mathrm{g}\) of \(\mathrm{B}_{2} \mathrm{O}_{3}\) when burned in excess \(\mathrm{O}_{2},\) what is its empirical formula?
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