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
Final answer is molarity of the H2SO4 btw.
What makes you think it's wrong?
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.
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.
Didn't work :( going to have to ask my teacher about this one, because im almost 100% positive its correct.
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.
Your calculations are correct up to this point. However, you made a mistake in the final step of your calculation.
To determine the molarity of the sulfuric acid solution, you need to divide the moles of sulfuric acid by the volume of the sulfuric acid solution.
In your calculation, you correctly determined that the moles of NaOH is 0.001272 mol. However, since the balanced equation shows a 1:2 ratio between NaOH and H2SO4, you need to multiply the moles of NaOH by 0.5 (or divide by 2) to get the moles of H2SO4:
0.001272 mol NaOH x (0.5 mol H2SO4/1 mol NaOH) = 0.000636 mol H2SO4
Next, divide the moles of H2SO4 by the volume of the sulfuric acid solution in liters:
0.000636 mol H2SO4 / 0.043 L H2SO4 = 0.014791 mol/L
So, the correct molarity of the sulfuric acid solution is 0.014791 mol/L.