Question

Calculate the new molarity that results when \(250 . \mathrm{mL}\) of water is added to each of the following solutions. a. \(125 \mathrm{~mL}\) of \(0.251 \mathrm{M} \mathrm{HCl}\) b. \(445 \mathrm{~mL}\) of \(0.499 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) c. \(5.25 \mathrm{~L}\) of \(0.101 \mathrm{M} \mathrm{HNO}_{3}\) d. \(11.2 \mathrm{~mL}\) of \(14.5 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short answer

The new molarity for each solution is as follows: a) 0.0837 M, b) 0.319 M, c) 0.0969 M, and d) 0.619 M.

Step-by-step solution

a. Calculate the new molarity of HCl solution#

First, let's write down the given values: Initial molarity, \(M_1 = 0.251 M\) Initial volume, \(V_1 = 125 mL\) Volume of water added, \(V_{water} = 250 mL\) To find the final volume, add the initial volume and the volume of water added: \(V_2 = V_1 + V_{water} = 125 mL + 250 mL = 375 mL\) Now, use the formula \(M_1V_1 = M_2V_2\) to find the final molarity: \(0.251 M \cdot 125 mL = M_2 \cdot 375 mL\) \(M_2 = \frac{0.251 M \cdot 125 mL}{375 mL} = 0.0837 M\) So, the new molarity of the HCl solution is \(0.0837 M\).

b. Calculate the new molarity of H2SO4 solution#

Initial molarity, \(M_1 = 0.499 M\) Initial volume, \(V_1 = 445 mL\) Volume of water added, \(V_{water} = 250 mL\) Final volume, \(V_2 = V_1 + V_{water} = 445 mL + 250 mL = 695 mL\) Using the formula \(M_1V_1 = M_2V_2\) : \(0.499 M \cdot 445 mL = M_2 \cdot 695 mL\) \(M_2 = \frac{0.499 M \cdot 445 mL}{695 mL} = 0.319 M\) The new molarity of the H2SO4 solution is \(0.319 M\).

c. Calculate the new molarity of HNO3 solution#

Initial molarity, \(M_1 = 0.101 M\) Initial volume, \(V_1 = 5.25 L\) We need to convert this to mL: \(V_1 = 5.25 L \cdot 1000 mL/L = 5250 mL\) Volume of water added, \(V_{water} = 250 mL\) Final volume, \(V_2 = V_1 + V_{water} = 5250 mL + 250 mL = 5500 mL\) Using the formula \(M_1V_1 = M_2V_2\): \(0.101 M \cdot 5250 mL = M_2 \cdot 5500 mL\) \(M_2 = \frac{0.101 M \cdot 5250 mL}{5500 mL} = 0.0969 M\) The new molarity of the HNO3 solution is \(0.0969 M\).

d. Calculate the new molarity of HC2H3O2 solution#

Initial molarity, \(M_1 = 14.5 M\) Initial volume, \(V_1 = 11.2 mL\) Volume of water added, \(V_{water} = 250 mL\) Final volume, \(V_2 = V_1 + V_{water} = 11.2 mL + 250 mL = 261.2 mL\) Using the formula \(M_1V_1 = M_2V_2\): \(14.5 M \cdot 11.2 mL = M_2 \cdot 261.2 mL\) \(M_2 = \frac{14.5 M \cdot 11.2 mL}{261.2 mL} = 0.619 M\) The new molarity of the HC2H3O2 solution is \(0.619 M\).

Key concepts

Understanding Molarity

Molarity, often denoted by 'M', is a fundamental concept in chemistry that quantifies the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. In mathematical form, the molarity (\(M\)) can be expressed as: \[ M = \frac{\mathrm{moles\ of\ solute}}{\mathrm{volume\ of\ solution\ in\ liters}} \]

Molarity is a critical component when it comes to mixing chemical solutions because it allows chemists to predict the outcome of reactions and the behavior of ions in solution. When calculating molarity, it's essential to remember to convert all volumes to the same unit, typically liters for standardization. For students grappling with molarity, a useful tip is to carefully track the units involved in the calculation, as this can prevent common errors related to volume and concentration measurements.

Solution Dilution Explained

The concept of solution dilution involves adding more solvent (often water) to a solute to decrease the concentration of the solution. The process does not change the amount of solute present; it simply disperses it within a greater volume of solvent. One might wonder why this is helpful. Well, dilution is a fundamental technique in laboratory practices to achieve the desired concentration for various applications.

To understand dilution better, consider the formula: \[ M_1V_1 = M_2V_2 \]
This equation represents the relationship between the initial and final concentrations (\(M_1, M_2\) respectively) and volumes (\(V_1, V_2\) respectively) of a solution. To improve comprehension, students should solve dilution problems by carefully listing known and unknown variables and using the equation to solve for the unknown molarity after dilution. Emphasizing that the product of concentration and volume before and after dilution remains constant can also help students in grasping this concept.

Molar Concentration in Practice

Molar concentration, commonly referred to as molarity, indicates how concentrated a solution is with respect to the amount of solute present per unit volume of solution. It helps chemists standardize reactions and ensures that the quantities of reactants are appropriate for the intended reaction.

In practice, when we speak of increasing or decreasing molar concentration, we are typically adding or removing a solvent or combining solutions with different molarities. When performing molarity calculations, it is important always to cross-check the units to prevent errors and achieve accurate results. As a further tip, always ensure that the solute's molecular weight and the solution's volume are correctly used to find moles, which is a step toward calculating the molarity. This attention to detail ensures that students can reliably execute experiments and understand the proportional relationship between the number of moles of a compound and the volume of the solution.