Solutions, dilution and titration in code

Molarity in code

Molarity is the number of moles of solute in one litre of solution, in mol/L (often written M). In the laboratory, volumes are in millilitres, so a function must divide by $1000$:

• moles $=$ molarity $\times \dfrac{\text{volume in mL}}{1000}$

Making a solution

To make a solution of a given molarity, weigh the solute, dissolve it, and make it up to the mark in a volumetric flask. The mass to weigh is the moles times the molar mass:

• mass $=$ molarity $\times \dfrac{V}{1000} \times M$

Dilution and mixing

Adding water does not change the moles of solute, only the volume. So:

• $M_1 V_1 = M_2 V_2$, and the volume of stock to take is $V_1 = \dfrac{M_2 V_2}{M_1}$.

When two solutions of the same solute are mixed, the moles add up and the volumes add up:

• $M = \dfrac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$

The end point of a titration

In a titration, a base runs from a burette into an acid until the acid is just used up. At the end point, the moles match the ratio in the equation. For hydrochloric acid and sodium hydroxide the ratio is $1 : 1$. For sulphuric acid, $\text{H}_2\text{SO}_4$, each mole needs two moles of base.

• volume of base at the end point $= \dfrac{M_{\text{acid}} \times V_{\text{acid}} \times \text{ratio}}{M_{\text{base}}}$

A function with a default argument, ratio=1, covers both cases: leave it out for $1 : 1$, or give ratio=2.

Practise with new numbers

Try it in Python

The program has three small functions, then a loop that follows the acid left in the flask as the base runs in.

Common mistakes

• Forgetting to divide by 1000. A volume in millilitres gives moles a thousand times too large.
• Solving $M_1V_1 = M_2V_2$ the wrong way round. The stock is more concentrated, so its volume is the smaller one.
• Adding molarities when mixing. Add the moles, then divide by the total volume.
• Leaving out the ratio. Sulphuric acid needs twice the base of hydrochloric acid at the same molarity.
• Calling with the wrong number of values. end_point(0.1, 20) stops with TypeError: end_point() missing 1 required positional argument: 'base_M'.

Remember these

• Moles $=$ molarity $\times \dfrac{V \text{ in mL}}{1000}$; mass to weigh $=$ moles $\times M$.
• Dilution keeps the moles: $M_1V_1 = M_2V_2$.
• Mixing: add the moles and add the volumes.
• End point: the moles of acid and base match the ratio in the equation; a default argument such as ratio=1 keeps the common case short.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What does moles_in(0.25, 200) return?

2. You dilute $10\ \text{mL}$ of a $2\ \text{mol/L}$ solution to $100\ \text{mL}$. What is the new molarity?

3. You mix $100\ \text{mL}$ of $0.3\ \text{mol/L}$ salt water with $100\ \text{mL}$ of $0.1\ \text{mol/L}$ salt water. What is the molarity of the mixture?

4. Why does the function end_point have ratio=1 in its first line?

5. In the loop of the program, why is the acid left written as max(left, 0)?