Answer:
0.19 M is the concentration of H₂SO₄ that remains after neutralization
Explanation:
In a neutralization reaction we produce water and a salt, formed by the cation and anion from the correspond base and correspond acid.
Acid: H₂SO₄
Base: KOH
We determine the reaction: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
1 mol of sulfuric reacts with 2 moles of hydroxide to produce 1 mol of sulfate and 2 moles of water.
We determine the moles of each reactant ( M . volume)
0.650L . 0.490mol/L = 0.32 moles of acid
0.600L . 0.280mol/L = 0.17 moles of base
In this question: "What concentration of sulfuric acid remains after neutralization" it is shown that the excess reactant is the H₂SO₄
We verify that, 2 moles of base can react with 1 mol of acid
So, 0.17 moles of base must react with (0.17 .1) / 2 = 0.085 moles
Moles of acid that remains, after the reaction is complete: 0.32 - 0.085 =
0.235 moles.
To determine the concentration, our total volume is (base vol + acid vol)
0.650L + 0.600L = 1.250L → M (mol/L) = 0.235 mol / 1.250L = 0.19 M
