Ask questions and get helpful responses.

chemistry

A 43 mL sample of a solution of sulfuric acid is neutralized by 24 mL of a 0.053 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution.
Answer in units of mol/L.

Here is my work so far. I thought I have successfully solved the problem, but apparently my answer Is wrong.

H2SO4 + 2NaOH = 2H2O + Na2SO4

.024L NaOH x .053M NaOH = .001272 mol NaOH
.001272 mol NaOH x (1 mol H2SO4/2 mol NaOH) / .043L H2SO4
= .0147906977

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
  1. Final answer is molarity of the H2SO4 btw.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. What makes you think it's wrong?

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. The website we use for our homework tells you when your answer is wrong right when you type it in, and mine is apparently wrong.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  4. You answer of 0,0147906,,,,, should be rounded to 0.015, You may not have rounded. You may not have added M or mols/L ater the 0.015. Most of these auto checking websites are unforgiving on significant figures. Try the 0.015 and if that doesn't work bry 0.015 M.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  5. Didn't work :( going to have to ask my teacher about this one, because im almost 100% positive its correct.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  6. Try 0.030 as an answer. Maybe the website wants to know what the polarity is if you were just titrating to obtain the first equivalent point. I agree with 0.015, but 0.030 is worth a shot.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Respond to this Question

First Name

Your Response

Still need help? You can ask a new question.