In my latest chem lab the objective was to create a primary standard of $\ce{NaOH}$ and use it to determine the concentration of sulfuric acid.

The first part of the lab was determine the molarity of the $\ce{NaOH}$ solution through a series of titrations.

  • A sample of KHP (abbreviated form of $\ce{KHC8H4O4}$) was placed into a flask with approximately $\pu{25 ml}$ of water.

  • Phenolphthalein was added to the flask as the indicator. $\ce{NaOH}$ was then titrated into the flask with a burette. From multiple titrations of this sort I was able to calculate the molarity of $\ce{NaOH}$.

Below I have included part of my table and calculations (Note: $\pu{1 mol}$ of KHP is equal to $\pu{1 mol}$ of $\ce{NaOH}$ in this experiment. If I have made any mistakes please tell me).

Trial 1:
Mass of KHP in flask $= \pu{0.5108 g}$
Volume of $\ce{NaOH}$ added to flask $ = \pu{21.73 ml}$
Calculation of molarity of $\ce{NaOH}$ for trial 1:
Molar mass of KHP $ = \pu{204.23 g/mol}$
$\pu{0.5108 g}/\pu{204.23 g/mol} = \pu{0.002501 mol}$ of KHP, which is equal to $\pu{0.002501 mol}$ of $\ce{NaOH}$.
Molarity of $\ce{NaOH} = \pu{0.002501 mol}/\pu{0.02173 L} = \pu{0.1151 M}$

I did 3 other trials like this (in total 4) and calculated the average molarity of $\ce{NaOH}$ to be $\pu{0.1159 M}$.

The second half of the lab is the part I had trouble with.
We were given a sample of $\ce{H2SO4}$ with an unknown concentration. I took $\pu{10 ml}$ of this $\ce{H2SO4}$ and mixed it with $\pu{100 ml}$ of distilled water. This new diluted solution of $\ce{H2SO4}$ (I will refer to it as solution 2 now) was the solution used in the trials to determine the molarity. So $\pu{25 ml}$ of solution 2 was added to a flask with a few drops of phenolphthalein. A titration using $\ce{NaOH}$ (the same $\ce{NaOH}$ as used in the previous section) was performed.

My task is to now figure out the concentration of the original $\ce{H2SO4}$ solution. I have tried 2 different methods. The first method I attempted seems so flawed I didn't bother to put it on (it didn't even make sense to me). Each method seems incorrect and have yielded drastically different results. Below I have provided a sample of my table and one of my attempts to solve for the molarity of $\ce{H2SO4}$.

The net ionic equation of this procedure is: $$\ce{H2SO4 +2NaOH <=> Na2SO4 + 2H2O}$$

Trial 1: Volume of diluted acid (solution 2) in flask: $\pu{25.00 ml}$
Volume of $\ce{NaOH}$ added to flask: $\pu{23.81 ml}$

Attempt 1 at finding molarity:
Moles of $\ce{NaOH}$ added to flask: $\pu{0.02381 L} \cdot \pu{0.1159 M} = \pu{0.0027596 mol}$ $\ce{NaOH}$

Amount of of $\ce{H2SO4}$: $0.0027596/2 = \pu{0.0013798 mol}$ $\ce{H2SO4}$ (The 2 came from the net ionic equation above)
Molarity of diluted $\ce{H2SO4}$ (solution 2): $\pu{0.0013798 mol}/ \pu{0.025 L}= \pu{0.054172 M}$
(I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?)

\begin{align} C_1V_1 &= C_2V_2\\ C_1&=?\\ V_1&= \pu{0.01 L}\\ C_2&= \pu{0.054172 M}\\ V_2&=\pu{0.1 L}\\ \text{Therefore:}\\ C_1&=(\pu{0.054172 M})\cdot(\pu{0.1 L})/(\pu{0.01 L})\\ C_1&=\pu{0.54172 M}\\ \end{align} Molarity of original/stock $\ce{H2SO4}$.

  • $\begingroup$ You were supposed to measure 10ml of the acid in a 100ml capacity volumetric flask and make it up to mark with ditilled water. That is 10ml acid with 90ml distilled water. This will give you a dilution factor of 10. $\endgroup$ Commented Dec 6, 2020 at 13:09

1 Answer 1


Molarity of diluted $\ce {H2SO4}$ (solution 2): $\pu{0.0013798 mol}/\pu{0.025 L} = \pu{0.054172 M}$
(I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?)

No, you used the correct volume since you want to know the concentration of the $\pu{25 mL}$ that you added to the flask.

$\pu{0.54172 M}$ - Molarity of original solution of $\ce {H2SO4}$.


  • $\begingroup$ "I took $\pu{10 ml}$ of this $\ce{H2SO4}$ and mixed it with $\pu{100 ml}$ of distilled water" Is $V_2=\pu{0.11 L}$ or $ V_2=\pu{0.1 L}$ $\endgroup$ Commented Dec 8, 2019 at 1:53
  • 1
    $\begingroup$ @AdnanAL-Amleh If you take 10 mL of one solution and add 100 mL of another solution, you will always have more than 100 mL. $\endgroup$
    – LDC3
    Commented Dec 9, 2019 at 2:18

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