Question

How many milliliters of each of the following solutions will provide \(25.0 \mathrm{~g}\) of \(\mathrm{KOH}\) ? a. \(2.50 \mathrm{M} \mathrm{KOH}\) solution b. \(0.750 \mathrm{M}\) KOH solution c. \(5.60 \mathrm{M} \mathrm{KOH}\) solution

Short answer

a. 178 mL b. 593 mL c. 79.5 mL

Step-by-step solution

Calculate the moles of \text{KOH}

Use the molar mass of \text{KOH} to convert grams to moles. The molar mass of \text{KOH} is approximately 56.11 \text{g/mol}. \[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{25.0 \text{ g}}{56.11 \text{ g/mol}} \ n = 0.445 \text{ mol} \]

Using Molarity Equation for Part (a) - 2.50 M KOH solution

The molarity equation is given by \[ M = \frac{n}{V} \] Rearrange it to solve for volume: \[ V = \frac{n}{M} = \frac{0.445 \text{ mol}}{2.50 \text{ M}} \ V = 0.178 \text{ L} \ V = 178 \text{ mL} \]

Using Molarity Equation for Part (b) - 0.750 M KOH solution

Using the same equation: \[ V = \frac{n}{M} = \frac{0.445 \text{ mol}}{0.750 \text{ M}} \ V = 0.593 \text{ L} \ V = 593 \text{ mL} \]

Using Molarity Equation for Part (c) - 5.60 M KOH solution

Using the same equation: \[ V = \frac{n}{M} = \frac{0.445 \text{ mol}}{5.60 \text{ M}} \ V = 0.0795 \text{ L} \ V = 79.5 \text{ mL} \]

Key concepts

molarity

Molarity is a way to measure the concentration of a solution. It tells you how many moles of a solute (the substance being dissolved) are present in one liter of solution. The formula for molarity is:
\( M = \frac{n}{V} \)
where \( M \) is the molarity, \( n \) is the number of moles of the solute, and \( V \) is the volume of the solution in liters.

To understand molarity better, think about making lemonade. If you add more lemon juice (solute) to the same amount of water (solvent), the lemonade gets stronger (more concentrated). Molarity helps you quantify that concentration, making it useful in chemistry for precise measurements.
In the exercise, we used the molarity formula to find out how many milliliters of different KOH solutions are needed to get 25 grams of KOH.

moles and molar mass

A mole is a unit that helps chemists count particles like atoms or molecules. One mole is \( 6.022 \times 10^{23} \) particles, known as Avogadro's number. Molar mass, on the other hand, is the mass of one mole of a substance. For example, the molar mass of potassium hydroxide (KOH) is approximately 56.11 g/mol.

To convert grams to moles, you use the formula:
\( n = \frac{mass}{molar mass} \)
In the exercise, we needed to convert 25 grams of KOH to moles:
\( n = \frac{25.0 \text{ g}}{56.11 \text{ g/mol}} = 0.445 \text{ mol} \).
This gave us the number of moles of KOH required to proceed with the calculations.

solution concentration calculation

Now, let's talk about calculating solution concentration. Once you know the number of moles (\( n \)) and the molarity (\( M \)), you can find out the volume (\( V \)) of the solution needed. You can rearrange the molarity formula to solve for volume:
\( V = \frac{n}{M} \).

Part (a): For a 2.50 M KOH solution, \( V = \frac{0.445 \text{ mol}}{2.50 \text{ M}} = 0.178 \text{ L} = 178 \text{ mL} \).
Part (b): For a 0.750 M KOH solution, \( V = \frac{0.445 \text{ mol}}{0.750 \text{ M}} = 0.593 \text{ L} = 593 \text{ mL} \).
Part (c): For a 5.60 M KOH solution, \( V = \frac{0.445 \text{ mol}}{5.60 \text{ M}} = 0.0795 \text{ L} = 79.5 \text{ mL} \).

These calculations show how different concentrations of the same solute require different volumes to provide the same number of moles.