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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 $\ce{H2SO4}$.

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

No, you used the correct volume since you want to know the concentration of the 25 mL that you added to the flask.

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

Correct.

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