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Chapter 15: Problem 61

When \(250.0 \mathrm{~mL}\) of water are added to \(10.0 \mathrm{~mL}\) of \(12 \mathrm{M} \mathrm{HCl}\), what will be the new molarity?

Short Answer

Expert verified
The new molarity of the solution is approximately \(0.46 \text{M}\).

Step by step solution

01

- Calculate Initial Moles of HCl

First, calculate the number of moles of \(\text{HCl}\) in the 10.0 mL solution. Use the formula \( \text{moles} = \text{molarity} \times \text{volume in liters} \). \(\text{moles of HCl} = 12 \text{M} \times 0.010 \text{L} = 0.12 \text{moles}\)
02

- Determine Total Volume of Solution

Add the volumes of the \(12 \text{M HCl}\) solution and the water. The total volume: \(10.0 \text{mL} + 250.0 \text{mL} = 260.0 \text{mL} \), which is \(0.260 \text{L}\).
03

- Calculate New Molarity

Use the formula for molarity: \( \text{molarity} = \frac{\text{moles of solute}}{\text{total volume of solution in liters}} \). \(\text{New molarity} = \frac{0.12 \text{moles}}{0.260 \text{L}} \approx 0.46 \text{M}\)

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Key Concepts

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

Moles of Solute
To understand molarity calculation, we must first understand the concept of moles of solute. The 'solute' is the substance dissolved in a solvent to form a solution.
The 'mole' is a standard unit in chemistry that measures the amount of a substance. It links the mass of the substance to its number of particles.
When calculating moles, we use the formula: \( \text{moles} = \text{molarity} \times \text{volume in liters} \).
This essentially tells us how many moles of the solute (e.g., HCl) are present in a certain volume of solution.
Solution Volume
The total volume of a solution is crucial in molarity calculations. It is the sum of the volumes of the solute and the solvent. In our exercise, the initial volume (10.0 mL of HCl) is combined with the added volume (250.0 mL of water).
To ensure accurate calculations, always convert the volume to liters: \(250.0 \text{ mL} + 10.0 \text{ mL} = 260.0 \text{ mL} = 0.260 \text{ L} \).
Understanding this ensures we can properly use the molarity formula.
Dilution Calculation
Dilution involves reducing the concentration of a solute in a solution, usually by adding more solvent. When we dilute a solution, the number of moles of solute remains constant, but the total volume increases.
After calculating the initial moles of solute and the total volume of the new solution, we find the new molarity using the formula: \( \text{molarity} = \frac{\text{moles of solute}}{\text{total volume of solution in liters}} \).
In our exercise, \( \text{New molarity} = \frac{0.12 \text{ moles}}{0.260 \text{ L}} ≈ 0.46 \text{ M} \).
This step ensures we understand the concentration change due to dilution.

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Most popular questions from this chapter

For each of the given pairs, determine which solution is more acidic. All are water solutions. Explain your answer. (a) \(1 \mathrm{M} \mathrm{HCl}\) or \(2 \mathrm{M} \mathrm{HCl}\) ? (b) \(1 M \mathrm{HNO}_{3}\) or \(1 M \mathrm{H}_{2} \mathrm{SO}_{4}\) ?

Write an equation to illustrate the acid-base reaction that will take place between perchloric acid, \(\mathrm{HClO}_{4}\), and a hydroxide ion, \(\mathrm{OH}^{-}\). Identify the acids, bases, and their conjugates.

What volume (mL) of \(0.4233 M \mathrm{H}_{2} \mathrm{SO}_{4}\) is needed to neutralize \(6.38 \mathrm{~g} \mathrm{KOH}\) ?

Determine how many grams of \(\mathrm{Al}(\mathrm{OH})_{3}\) will be required to neutralize \(275 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{HCl}\) according to the reaction: $$ 3 \mathrm{HCl}(a q)+\mathrm{Al}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$

A batch of sodium hydroxide was found to contain sodium chloride as an impurity. To determine the amount of impurity, a \(1.00\)-g sample was analyzed and found to require \(49.90 \mathrm{~mL}\) of \(0.466 \mathrm{M}\) \(\mathrm{HCl}\) for neutralization. What is the percent of \(\mathrm{NaCl}\) in the sample?
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