4.3.4 - Molar concentration and titration calculations

Molar concentration, amount and volume (Higher)

Molar concentration states the amount of solute in moles per unit volume of the complete solution.

A solution at $0.200\,\mathrm{mol\,dm}^{-3}$ contains 0.200 mol of solute per cubic decimetre. It does not contain 0.200 mol in every sample regardless of sample size. The solute is the substance dissolved; use the final solution volume, not just the volume of solvent added.

For $25.0\,\mathrm{cm}^{3}$ solution at $0.200\,\mathrm{mol\,dm}^{-3}$, first convert cubic centimetres to cubic decimetres:

The sample is 1/40 of a cubic decimetre, so its solute amount is 1/40 of 0.200 mol.

To find concentration from $0.0150\,\mathrm{mol}$ solute in $0.100\,\mathrm{dm}^{3}$:

The second line checks the scale of the concentration. To find the volume containing $0.0300\,\mathrm{mol}$ at this concentration, use $V=n/c=0.0300/0.150=0.200\,\mathrm{dm}^{3}$, or $200\,\mathrm{cm}^{3}$. Concentration times volume gives moles; moles divided by concentration gives volume.

Linking mass and molar concentration (Higher)

Molar mass connects concentration in moles per cubic decimetre with concentration in grams per cubic decimetre.

Since $m=nM$ and $n=cV$, solute mass is $m=cVM$ when molar mass $M$ is in $\mathrm{g\,mol}^{-1}$. For $0.250\,\mathrm{dm}^{3}$ sodium hydroxide solution at $0.200\,\mathrm{mol\,dm}^{-3}$, with $M_r(\ce{NaOH})=40$:

For the same solute:

Thus $0.200\,\mathrm{mol\,dm}^{-3}$ sodium hydroxide corresponds to $0.200\times40=8.00\,\mathrm{g\,dm}^{-3}$. In the reverse direction, divide mass concentration by molar mass. Different substances at the same molar concentration can have different mass concentrations because their particles have different masses.

To find molar concentration from a preparation, calculate amount before dividing by volume. For 4.00 g sodium hydroxide made up to a final solution volume of $250\,\mathrm{cm}^{3}$:

This is equivalent to $c=m/(MV)$. Doubling the solute mass while keeping the same final volume has doubled the concentration relative to the first example.

For a fixed solute and fixed solution volume, doubling mass doubles both amount and molar concentration. At fixed mass, doubling solution volume halves concentration. Dividing a well-mixed solution into two samples halves both mass and volume, leaving concentration unchanged. Dilution by adding solvent keeps the solute amount but increases volume, so concentration falls.

Mole ratios in a titration (Higher)

A titration finds the volume of one solution needed to react exactly with a measured volume of another solution.

A volumetric pipette transfers a known volume into a conical flask and a burette measures the added solution. A suitable indicator gives a clear end point; adding drops near it and swirling helps avoid overshooting. Rinse the pipette with its transferred solution and the burette with its added solution to avoid dilution by residual water. Read the bottom of the meniscus at eye level. These measurements provide the reacting volumes for the calculation.

The titre is the final burette reading minus the initial reading, not the final reading alone. For initial reading $1.20\,\mathrm{cm}^{3}$ and final reading $21.40\,\mathrm{cm}^{3}$, the titre is $20.20\,\mathrm{cm}^{3}$. Concentration calculations use the titre and the pipetted volume.