# 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 $$\ce{NaOH}$$ 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 $$\ce{H2SO4}$$.

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 $$\ce {H2SO4}$$.
• "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 $V_2=\pu{0.1 L}$ – Adnan AL-Amleh Dec 8 '19 at 1:53