A student makes a standard solution of sulfuric acid by taking 10.00 mL of a super-concentrated stock solution and diluting it to 5.00 L. He then standardizes the diluted solution by titrating 30.00 mL with 34.68 mL of a 0.2458 M solution of KOH. What is the concentration of the stock solution and the standard solution?

Standard solution: 0.1421M

Stock solution: 71.04M

Explanation:

The reaction of H₂SO₄ with KOH produces:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

In titration, moles of KOH are:

0.03468L ₓ (0.2458mol / L) = 0.008524 moles of KOH. As 1 mole of acid reacts with 2 moles of KOH, moles of titrated acid are: 0.008524/2 = 0.004262 moles of H₂SO₄.

Thus, concentration of standard solution is:

0.004262 moles of H₂SO₄ / 0.03000L = 0.1421M standard solution

Now, standard solution was prepared as a dilution of 10.00mL to 5.00L of stock solution. That means concentration of stock solution is:

0.1421M × (5000mL / 10.00mL) = 71.04M stock solution

explanation: no but i do not usually comment