Question

Question 8 of 8> 0 Attempt 2 A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. He has a 0.1284 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard solution. The endpoint was reached after the addition of 19.15 mL of the standard solution. What is the concentration of the original sulfuric acid solution? concentration:025

Answer 1

12.29 M

Step-by-step explanation:

• The reaction that takes place is:

H₂SO₄ + 2NaOH → 2Na⁺ + SO₄⁻² + 2H₂O

• Now let's calculate the moles of H₂SO₄ that were titrated:

`0.1284 M * 0.1915L*(1molAcid)/(2molNaOH)` = 0.01229 mol H₂SO₄.

• Thus, the concentration of the diluted solution is:

0.01229 mol H₂SO₄ / 0.010 L = 1.229 M

• Finally, the concentration of the original acid solution is:

`1.229 M*(250mL)/(25mL)` = 12.29 M