Question

Calculate the new molarity when \(150 .\) mL of water is added to each of the following solutions. a. \(125 \mathrm{mL}\) of \(0.200 \mathrm{M} \mathrm{HBr}\) b. \(155 \mathrm{mL}\) of \(0.250 \mathrm{M} \mathrm{Ca}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\) c. 0.500 L of \(0.250 M \mathrm{H}_{3} \mathrm{PO}_{4}\) d. \(15 \mathrm{mL}\) of \(18.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\)

Short answer

a. The new molarity of the HBr solution is approximately \(0.091 M\). b. The new molarity of the Ca(C2H3O2)2 solution is approximately \(0.127 M\). c. The new molarity of the H3PO4 solution is approximately \(0.192 M\). d. The new molarity of the H2SO4 solution is approximately \(1.636 M\).

Step-by-step solution

(Solution for part a)

: 1. Identify initial volume and molarity: V1 = 125 mL and C1 = 0.200 M. 2. Calculate the total volume after adding water: V2 = V1 + 150 mL = 125 mL + 150 mL = 275 mL. 3. Apply the dilution formula: C1V1 = C2V2. Rearrange the formula to find the new molarity: C2 = C1V1 / V2. 4. Plug in the values: C2 = (0.200 M)(125 mL) / (275 mL). 5. Calculate the new molarity: C2 = 0.091 M (approximately).

(Solution for part b)

: 1. Identify initial volume and molarity: V1 = 155 mL and C1 = 0.250 M. 2. Calculate the total volume after adding water: V2 = V1 + 150 mL = 155 mL + 150 mL = 305 mL. 3. Apply the dilution formula: C1V1 = C2V2. Rearrange the formula to find the new molarity: C2 = C1V1 / V2. 4. Plug in the values: C2 = (0.250 M)(155 mL) / (305 mL). 5. Calculate the new molarity: C2 = 0.127 M (approximately).

(Solution for part c)

: 1. Identify initial volume and molarity: V1 = 0.500 L and C1 = 0.250 M. 2. Calculate the total volume after adding water: V2 = V1 + 0.150 L = 0.500 L + 0.150 L = 0.650 L. (Note: converted 150 mL to 0.150 L) 3. Apply the dilution formula: C1V1 = C2V2. Rearrange the formula to find the new molarity: C2 = C1V1 / V2. 4. Plug in the values: C2 = (0.250 M)(0.500 L) / (0.650 L). 5. Calculate the new molarity: C2 = 0.192 M (approximately).

(Solution for part d)

: 1. Identify initial volume and molarity: V1 = 15 mL and C1 = 18.0 M. 2. Calculate the total volume after adding water: V2 = V1 + 150 mL = 15 mL + 150 mL = 165 mL. 3. Apply the dilution formula: C1V1 = C2V2. Rearrange the formula to find the new molarity: C2 = C1V1 / V2. 4. Plug in the values: C2 = (18.0 M)(15 mL) / (165 mL). 5. Calculate the new molarity: C2 = 1.636 M (approximately).

Key concepts

Molarity

Molarity is a way to express the concentration of a chemical solution. It's expressed in moles per liter (mol/L), which tells you how many moles of a solute are present in one liter of solution. In the dilution exercises given, molarity is used to determine how concentrated a solution is before and after dilution.
For example, if you have a solution of hydrohalic acid (HBr) with a molarity of 0.200 M, this means there are 0.200 moles of HBr in every liter of solution. Knowing the molarity helps us understand the strength or potency of a solution at a given time.
Calculating molarity is particularly useful in chemistry because it provides a way to quantitatively evaluate how much of a solute is present in a solution compared to the solution's total volume.
Understanding molarity allows chemists to prepare solutions with precise reactivities and concentrations, which is crucial in research, pharmaceuticals, and everyday laboratory work.

Volume and Concentration

In calculating dilutions, the volume of a solution is crucial. Volume tells us how much space the solution occupies, and when combined with molarity, it helps ascertain the overall concentration of the solution.
When water is added to a solution, it increases the overall volume but reduces the concentration of the solute within that solution. This process is called dilution.
For instance, if you start with a certain volume of concentrated solution, say 125 mL of 0.200 M HBr, and add 150 mL of water, the total volume becomes 275 mL. The concentration, which initially started as 0.200 M, will decrease because the same amount of solute is now dispersed in a larger volume.
Manipulating the volume is a simple yet effective way to adjust concentrations, making it easier for chemists to achieve the desired reactivity or solution strength for their experiments or applications.

Chemical Solutions

Chemical solutions are homogenous mixtures comprised of two parts: a solute and a solvent. The solute is the substance that is dissolved, while the solvent is the medium that dissolves the solute. In most solutions, the solvent is typically a liquid, often water, which we commonly encounter in dilution exercises.
In the context of the given exercise, when a solute like HBr (hydrobromic acid) is added to water, it forms an aqueous solution where the acid is evenly distributed throughout the solvent.
Chemical solutions are fundamental in chemistry because they allow for the uniform interaction of molecules, facilitating reactions and processes that wouldn't occur in heterogeneous mixtures. They also permit the measurement and standardization of solute concentrations, crucial for reproducibility in scientific research, pharmaceuticals, and various industrial applications.
Understanding how different factors such as temperature, pressure, and concentration affect a chemical solution is vital for predicting the behavior of substances in chemical reactions.

Equation Rearrangement

Equation rearrangement is a mathematical method that allows one to isolate and solve for a specific variable in an equation. It is a critical skill in chemistry, especially when dealing with dilution calculations and other computations involving concentrations.
The formula for dilution is:

• 
  \( C_1V_1 = C_2V_2 \)

This equation states that the product of the initial concentration and its corresponding volume equals the product of the final concentration and its total volume after dilution.
If you want to find the new molarity, you rearrange the equation to solve for \( C_2 \):

• 
  \( C_2 = \frac{C_1V_1}{V_2} \)

Rearranging equations necessitates a clear understanding of algebraic principles so that variables can be successfully isolated and calculated.
Effective rearrangement of equations enables chemists to solve various calculation problems accurately, whether in laboratories or industrial settings.