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Standard Solution: Solution of Known Concentration

What is Concentration and Why Does it Matter?

Before we dive into standard solutions, we need to understand what concentration means. Imagine you are making lemonade. If you add one spoon of sugar to a glass of water, it will be less sweet than if you add five spoons. The amount of sugar in the same amount of water is different. In chemistry, this ratio is called concentration.

Concentration tells us how much of a substance (the solute) is dissolved in a specific volume of another substance (the solvent, usually water). Scientists need to know exact concentrations to make reactions predictable and reproducible. A standard solution is like a perfectly calibrated measuring cup for chemicals.

Primary and Secondary Standard Solutions

Not all standard solutions are created equal. They are categorized based on how their concentration is determined.

Primary Standard Solutions are the gold standard. They are prepared by directly weighing a pure, stable, and easily weighable solid (called a primary standard) and dissolving it in a precise volume of solvent. The concentration is calculated directly from the mass and volume. Examples include sodium carbonate ($Na_2CO_3$) for acid-base titrations or potassium dichromate ($K_2Cr_2O_7$) for redox titrations.

Secondary Standard Solutions are solutions whose concentration is determined (or standardized) by titrating it against a primary standard solution. They are used when the solute isn't stable or pure enough to be a primary standard. A common example is sodium hydroxide ($NaOH$) solution, which absorbs water and carbon dioxide from air, making its exact mass unreliable. We first prepare an approximate concentration, then use a primary standard like oxalic acid to find its exact concentration.

Step-by-Step: Preparing a Standard Solution

Let's walk through the process of making a primary standard solution of sodium carbonate, $Na_2CO_3$, with a concentration of 0.100 M and a volume of 250.0 mL.

Step 1: Calculate the Required Mass. We use the molarity formula. First, find moles needed: $n = C \times V = 0.100 \text{ mol/L} \times 0.2500 \text{ L} = 0.02500 \text{ mol}$. The molar mass of $Na_2CO_3$ is about 106.0 g/mol. So, mass = moles × molar mass = $0.02500 \text{ mol} \times 106.0 \text{ g/mol} = 2.650 \text{ g}$.

Step 2: Weigh the Solid. Using an analytical balance, accurately weigh out exactly 2.650 g of pure, dry $Na_2CO_3$.

Step 3: Dissolve in a Beaker. Transfer the solid to a clean beaker. Add distilled water and stir gently until it completely dissolves.

Step 4: Transfer to a Volumetric Flask. Carefully pour the solution into a 250.0 mL volumetric flask. Rinse the beaker and stirrer with distilled water and add the rinsings to the flask to ensure all solute is transferred.

Step 5: Make Up to the Mark. Add distilled water to the flask until the bottom of the meniscus (the curved surface of the liquid) just touches the calibration mark on the flask's neck. Stopper and invert repeatedly to mix thoroughly. The solution's concentration is now precisely 0.100 M.

The Heart of Analysis: Titration in Action

Titration is the most common application of a standard solution. It is a technique to determine the concentration of an unknown solution by reacting it with a standard solution of known concentration.

Let's say we have a bottle of vinegar and want to know its acetic acid ($CH_3COOH$) concentration. We can use our standard 0.100 M $NaOH$ solution (which we previously standardized).

1. Setup: A known volume of vinegar (e.g., 10.00 mL) is placed in a flask with an indicator. The standard $NaOH$ solution is placed in a burette, a long glass tube with precise volume markings.

2. The Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$. The $NaOH$ is slowly added to the vinegar.

3. The Endpoint: The indicator changes color when all the acetic acid has just reacted with the sodium hydroxide. This is the equivalence point. We note the volume of $NaOH$ used from the burette, say 16.50 mL.

4. The Calculation:
Moles of $NaOH$ used: $n_{NaOH} = C_{NaOH} \times V_{NaOH} = 0.100 \text{ M} \times 0.01650 \text{ L} = 0.00165 \text{ mol}$.
From the reaction, 1 mole of $NaOH$ reacts with 1 mole of $CH_3COOH$. So, moles of $CH_3COOH$ in the vinegar sample = 0.00165 mol.
This was in 10.00 mL of vinegar. So, concentration of acetic acid:
$C_{acid} = \frac{0.00165 \text{ mol}}{0.01000 \text{ L}} = 0.165 \text{ M}$.

We have successfully used a standard solution to find an unknown concentration!

Important Questions

Footnote

^[1] Equivalents: In chemistry, an equivalent is the amount of a substance that reacts with or supplies one mole of hydrogen ions ($H^+$) in an acid-base reaction, or one mole of electrons in a redox reaction. Normality (N) is the number of equivalents per liter of solution.