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Volumetric - solution concentrations - titrations (acid - base)
Make a solution of known concentration - Primary standard
- We know the concentration because we know the mass, hence we know the moles.
Procedure to make primary standard
- Weigh solid (e.g. $Na_{2}CO_{3}$)
- Rinse with $H_{2}O$ to get all the sodium carbonate into the beaker!
- Mix with distilled water
- Transfer to volumetric flask
- Rinse the flask before with distilled water, not the solution cause it doesnt matter, all that matters is you know the moles of $Na_{2}CO_{3}$ inside
- Use a stirring rod? (optional :))
- DONT put your stirring rod anywhere, you will lose some of the mass to the blah blah blah, hence your calculations become more inaccurate!!! :(
- INSTEAD, wash the stirring rod with distilled water!
- Add distilled(should be self explanatory) water up to the mark, consider parallax error!!! oh no big scary
- If you goof up, and go above the mark, START AGAIN
- Transfer to storage bottle.
- RINSE THE BOTTLE WITH DISTILLED, YOU DON'T KNOW WHATS INSIDE!
- We will have some residual water left. We must rinse with the primary standard! This is so that the residual water after this will have the same concentration as the primary standard!!!
Primary Standard
Required:
- High purity
- Low affinity for water, i.e. not hygroscopic, deliquescent!
- Hygroscopic - pulls water from the air
- Deliquescent - pulls water and ends up in a slurry (a highly concentrated solution)
- Low affinity for air, as well, cause like we dont want our buddy to start combusting :/
- Classic example is $NaOH$.
Preferred:
- High molar mass
- This reduces the percentage of error in our calculations :)))
This will result in an accurately known solution.
- We will use this solution to standardise other solutions!!!!
Target:
- Essentially we reverse engineer the amount of mass we want to measure given we want it to be a specific concentration. This is just $n=cv$, $n=\frac{m}{M}$ backwards(solve for $m$).
Suppose we want $0.05molL^{-1}$, for a $250mL$ solution.
$$\begin{align}
n=cV, = 0.05 \times 0.250 = 0.0125mol \\
m = nM = 0.0125 \times (22.99 + 12.01 + 3 \times 16) \\
=1.04g
\end{align}
$$