Calculating Sulphuric Acid for Mixture Concentration

This problem involves calculating the amount of sulphuric acid needed to change the concentration of a mixture from an initial percentage to a target percentage. We start with a known volume of mixture and its initial concentration of sulphuric acid.

Let's break down the initial mixture:

• Total volume of mixture = 20 L
• Initial concentration of sulphuric acid = 10%
• Quantity of sulphuric acid initially present = 10% of 20 L
• Quantity of sulphuric acid = $\frac{10}{100} \times 20$ L = 2 L
• Quantity of water in the initial mixture = Total volume - Sulphuric acid quantity = 20 L - 2 L = 18 L

We want to add a certain quantity of sulphuric acid to this mixture. Let the quantity of sulphuric acid to be added be $x$ litres.

After adding $x$ litres of sulphuric acid:

• New total volume of the mixture = Initial total volume + Quantity of acid added = (20 + $x$) L
• New quantity of sulphuric acid in the mixture = Initial acid quantity + Quantity of acid added = (2 + $x$) L

The target concentration of sulphuric acid in the new mixture is 25%. The concentration is calculated as the ratio of the quantity of sulphuric acid to the total volume of the mixture, expressed as a percentage.

So, we can set up the following equation based on the target concentration:

$\text{New Percentage Concentration} = \left( \frac{\text{New Quantity of Sulphuric Acid}}{\text{New Total Volume}} \right) \times 100\%$

$25\% = \left( \frac{2 + x}{20 + x} \right) \times 100\%$

Now, we solve this equation for $x$ to find the quantity of sulphuric acid to be added.

Divide both sides by 100:

$\frac{25}{100} = \frac{2 + x}{20 + x}$

Simplify the fraction on the left side:

$\frac{1}{4} = \frac{2 + x}{20 + x}$

Cross-multiply:

$1 \times (20 + x) = 4 \times (2 + x)$

$20 + x = 8 + 4x$

Now, we need to isolate $x$. Subtract $x$ from both sides:

$20 = 8 + 4x - x$

$20 = 8 + 3x$

Subtract 8 from both sides:

$20 - 8 = 3x$

$12 = 3x$

Divide both sides by 3:

$x = \frac{12}{3}$

$x = 4$

So, 4 litres of sulphuric acid must be added to the mixture.

Let's check the result:

• Quantity of sulphuric acid added = 4 L
• New quantity of sulphuric acid = 2 L (initial) + 4 L (added) = 6 L
• New total volume = 20 L (initial) + 4 L (added) = 24 L
• New concentration = $\left( \frac{6}{24} \right) \times 100\% = \left( \frac{1}{4} \right) \times 100\% = 25\%$

The calculation confirms that adding 4 L of sulphuric acid results in a mixture with a 25% sulphuric acid concentration.

The quantity of sulphuric acid to be added is 4 L.

Revision Table: Mixture Problems

Mixture problems often involve calculating quantities or concentrations when substances are mixed or when a substance is added to a mixture. The key is often to track the amount of the specific component (like sulphuric acid here) and the total amount of the mixture.

Steps to solve mixture problems:

1. Identify the initial quantities and concentrations.
2. Identify what is being added or removed and its quantity/concentration.
3. Determine the effect of the change on the quantity of the component and the total quantity of the mixture.
4. Set up an equation based on the final concentration or desired ratio.
5. Solve the equation.

Additional Information: Concentration Units

Concentration can be expressed in various ways. In this problem, we used percentage by volume (volume of solute per total volume of solution). Other common concentration units include:

• Percentage by mass: Mass of solute per total mass of solution ($\frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%$).
• Molarity (M): Moles of solute per litre of solution ($\frac{\text{moles of solute}}{\text{litres of solution}}$).
• Molality (m): Moles of solute per kilogram of solvent ($\frac{\text{moles of solute}}{\text{kilograms of solvent}}$).
• Mole fraction: Moles of a component per total moles of all components in the mixture.

Understanding which unit is being used is crucial for setting up the correct equations in mixture problems.