# How do I calculate the concentration of sulphuric acid by a titration experiment with sodium hydroxide?

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 NaOH solution through a series of titrations.

• A sample of KHP (abbreviated form of $\ce{KHC8H4O4}$) was placed into a flask with approximately $25~\ce{ml}$ 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: 1 mole of KHP is equal to 1 mole of $\ce{NaOH}$ in this experiment. If I have made any mistakes please tell me.)

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

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

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 $10~\mathrm{ml}$ of this $\ce{H2SO4}$ and mixed it with $100~\mathrm{ml}$ 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 molarity. So $25~\mathrm{ml}$ 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 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: $25.00~\mathrm{mL}$
Volume of NaOH added to flask: $23.81~\mathrm{mL}$

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

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

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

Molarity of diluted $\ce {H2SO4}$ (solution 2): $0.0013798 mol/0.025 L=0.054172 M$
0.54172 M - Molarity of original solution of $\ce {H2SO4}$.