0.150 L of 0.490 M H2SO4 is mixed with 0.100 L of 0.250 M KOH. What concentration of sulfuric acid remains after neutralization?

Write the equation and balance it.

moles KOH = M x L
moles H2SO4 = M x L.
Remembering that it takes 2 moles KOH for every 1 mole H2SO4, determine which is in excess and subtract.

To find the concentration of sulfuric acid that remains after neutralization, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed.

First, let's calculate the number of moles of H2SO4 and KOH for the given volumes and concentrations:

For H2SO4:
Volume = 0.150 L
Concentration = 0.490 M
Number of moles of H2SO4 = Volume x Concentration
= 0.150 L x 0.490 mol/L
= 0.0735 mol

For KOH:
Volume = 0.100 L
Concentration = 0.250 M
Number of moles of KOH = Volume x Concentration
= 0.100 L x 0.250 mol/L
= 0.025 mol

The balanced equation for the reaction between H2SO4 and KOH is:

H2SO4 + 2KOH -> K2SO4 + 2H2O

From the equation, we can see that it takes one mole of H2SO4 to react with 2 moles of KOH.

Now, let's determine the limiting reactant:

To react completely with 0.0735 moles of H2SO4, we would need twice as many moles of KOH. So, the theoretical amount of KOH required is 2 x 0.0735 mol = 0.147 mol.

Since we have only 0.025 mol of KOH, it is the limiting reactant.

Now, let's calculate the remaining moles of H2SO4 after neutralization:

The amount of KOH used up is equal to the initial amount of KOH minus the remaining amount:

Remaining moles of KOH = Initial moles - Moles used
= 0.025 mol - 0.025 mol
= 0 mol

Since we have used up all of the KOH, the H2SO4 is in excess, which means the concentration of sulfuric acid remains the same.

Therefore, the concentration of sulfuric acid remains as 0.490 M.