Write the balanced neutralization reaction between H2SO4 and KOH in aq. solution.
.350 L of .410 M H2SO4 is mixed with .300 L of .230 M KOH. What concentration of sulfuric acid remains after neutralization?
______ M H2SO4
2KOH + H2SO4 ==> K2SO4 + 2H2O
Just another LR problem. You can ALWAYS spot LR problems because amounts for BOTH reactants are given in the problem. If just one reactants is given (and usually it say an excess of the other one is used) that is a simple stoichiometry problem like the Zn/HCl problem below.
The limiting reactant is the KOH with .069 moles right? So then I go from mol KOH to grams of K2SO4?
Yes, KOH is the limiting reagent. There's a new twist here. They don't ask for mols or grams of one of the products; instead they ask for how much H2SO4 is left. That's done the same way.
Convert mols KOH used (0.069) to mols H2SO4 used, subtract from H2SO4 there initially and that gives you mols H2SO4 left. Then mols H2SO4/L solution = M. Don't forget, total volume will be L KOH + L H2SO4.
.156 M?
Oh, Got it! .168?
I don't get either of those.
1.435 = mols H2SO4 initially
-0.069/2 = mols H2SO4 used
1.400 = mols H2SO4 left
Then M = mols/L = 1.400/0.65L = about 2.15M
To write the balanced neutralization reaction between H2SO4 and KOH, we need to consider the acid-base reaction that occurs between them. When an acid (H2SO4) reacts with a base (KOH), they undergo a neutralization reaction to form a salt (K2SO4) and water (H2O).
The balanced neutralization reaction between H2SO4 and KOH can be written as follows:
H2SO4 + 2KOH ⟶ K2SO4 + 2H2O
Now, let's calculate the moles of H2SO4 and KOH using the given volumes and concentrations:
For H2SO4:
Volume of H2SO4 = 0.350 L
Concentration of H2SO4 = 0.410 M
Moles of H2SO4 = Volume × Concentration = 0.350 L × 0.410 mol/L = 0.1435 mol
For KOH:
Volume of KOH = 0.300 L
Concentration of KOH = 0.230 M
Moles of KOH = Volume × Concentration = 0.300 L × 0.230 mol/L = 0.069 mol
Since H2SO4 and KOH react in a 1:2 ratio, we can see that the amount of KOH is limiting in this reaction. This means that all the KOH will react completely with H2SO4, and there will be some excess H2SO4 remaining after neutralization.
According to the balanced equation, 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, the number of moles of H2SO4 that will react is twice the number of moles of KOH, i.e., 2 × 0.069 mol = 0.138 mol.
To determine the concentration of H2SO4 that remains after neutralization, we need to calculate how much H2SO4 was not consumed in the reaction. To do this, subtract the moles of H2SO4 that reacted from the original moles of H2SO4:
Remaining moles of H2SO4 = Moles of H2SO4 - Moles of H2SO4 that reacted = 0.1435 mol - 0.138 mol = 0.0055 mol
Now, we can calculate the concentration of the remaining H2SO4:
Volume of H2SO4 remaining = Volume of H2SO4 initially = 0.350 L
Concentration of H2SO4 remaining = Remaining moles of H2SO4 / Volume of H2SO4 remaining
= 0.0055 mol / 0.350 L
Using these values, calculate the concentration of H2SO4 remaining after neutralization.