[Balrog]

Hi,

Can someone please help me with this problem?

Given the following neutralization reaction between sulphuric acid (H2SO4) and potassium hydroxide (KOH)

H2SO4 + 2KOH -> K2SO4 + 2H2O

If 0.150 L of 0.410 M H2SO4 is mixed with 0.100 L of 0.29O M KOH. What concentration of H2SO4 remains after neutralization?

Here is what I have done so far:

mol H2SO4 before neutralization is

0.150 L * 0.410 mol/L = 0.0615 mol H2SO4

mol KOH before neutralization is

0.100 L * 0.290 mol/L = 0.029 mol KOH

KOH appears to be our limiting reactant, so using stoichiometry we can say 1 mol H2SO4 reacts with 2 mol KOH, therefore 0.0290 mol KOH would react with

(0.0290 mol KOH * 1 mol H2SO4)/2 mol KOH = 0.0145 mol H2SO4

mol H2SO4 remaining after neutralization is

0.0615 mol - 0.0145 mol = 0.047 mol H2SO4

Then I use dimensional analysis to calculate the concentration of 0.047 mol H2SO4 that was left after neutralization as follows

(0.047 mol H2SO4 * 0.410 M H2SO4)/0.0615 mol H2SO4 = 0.313 M H2SO4

But 0.313 M H2SO4 is not the right answer. Where am I going wrong?

[Borek]

Then I use dimensional analysis to calculate the concentration of 0.047 mol H2SO4 that was left after neutralization as follows

(0.047 mol H2SO4 * 0.410 M H2SO4)/0.0615 mol H2SO4 = 0.313 M H2SO4

But 0.313 M H2SO4 is not the right answer. Where am I going wrong?

You were OK up to this moment. You known how many moles of the acid is left, calculate the volume, calculate the concentration. No idea what was the logic behind your last step, but it was incorrect.