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I just did a titration experiment at school, our teacher told us multiple times to add the solution of known concentration into the solution of unknown concentration. I am wondering if we can do it the other way, if not, why?

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From what I know, you can add the solution of unknown to the solution of known concentration. The only thing that matters is you have a known volume of the solution to which you're adding the other solution.

For example, suppose we're titrating $\pu{0.15 M}$ $\ce{Na2CO3}$ with an unknown concentration of $\ce{HCl}$, if we need to add $\pu{say 15 mL}$ $\ce{Na2CO3}$ to a known volume of say $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can calculate the concentration of the HCl solution.

$$\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}$$

$n(\ce{Na2CO3}) = \pu{0.15 mol/L} \cdot \pu{0.015 L}$

$n(\ce{Na2CO3}) = \pu{0.00225 mol}$ (i.e. the known concentration times 15 mL change to L)

We know that 2 moles of $\ce{HCl}$ will react with 1 mole of $\ce{Na2CO3}$ to form the products on the right hand side of the equation. So,

$n(\ce{HCl}) = 2 \cdot n(\ce{Na2CO3})$

which we calculated above using the relation of moles as product of concentration times volume. So

$n(\ce{HCl}) = \pu{0.0045 mol}$

Since we know that we had $\pu{30 mL}$ of a $\ce{HCl}$ solution with $\pu{0.0045 mol}$ of $\ce{HCl}$ in it,

$[\ce{HCl}] = \frac{\pu{0.0045 mol}}{\pu{0.03 L}}$

$[\ce{HCl}] = \pu{0.15 mol/L}$

So, we had a $\pu{0.15 M}$ solution of $\ce{HCl}$.

If we're now adding $\ce{HCl}$ to $\pu{15 mL}$ $\ce{Na2CO3}$ and it takes $\pu{30 mL}$ of $\ce{HCl}$ to reach the endpoint, we can use similar calculations to once again find the concentration of $\ce{HCl}$ (and get the same answer).

It doesn't matter whether you add the known solution to the unknown one, the only thing you need to know is the volume of the thing you're adding to and how much of the other one you need to add to reach the endpoint. Think of it like this, We have 4 variables so to speak; - concentration of the known solution (we know)

  • concentration of the unknown (what we want to work out)

  • volume of the known solution (either adding or added to) which we measure

  • volume of the unknown solution (either adding or added to)which we also measure

So we know 3 things no matter what we add what to and these 3 values are enough to work out the concentration of the unknown solution. You should clarify with your teacher what she meant. She may have said always add the acid to the base because if we place the base in the burette it may react with the glass, $\ce{SiO2}$ which acts as an acid and that might degrade up the precision for future use.

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    $\begingroup$ For future reference, let me suggest to include \mhchem while formatting your answers; functions enabled on ChemSE are compiled here: chemistry.meta.stackexchange.com/questions/3044/… As a second recommendation, I suggest to keep carrying (within reason) the units in the equations, which is eased with \pu{}. $\endgroup$ – Buttonwood Jun 4 '17 at 9:58
  • $\begingroup$ Thanks for your suggestions I appreciate it! I'm not too familiar with formatting thanks for directing me to that page. $\endgroup$ – shA3245699 Jun 4 '17 at 10:26

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