Write the balanced neutralization reaction that occurs between H 2 SO 4 and KOH in aqueous solution. Phases are optional. neutralization reaction: Suppose 0.150 L of 0.420 M H 2 SO 4 is mixed with 0.100 L of 0.210 M KOH . What concentration of sulfuric acid remains after neutralization

0.168 M

Explanation:

First, this is a reaction between the a strong base and a strong acid, therefore, we do not have to count with the acid constant of equilibrium. This reaction is taking place completely and occurs a neutralization, which is the following reaction:

H₂SO₄(aq) + 2KOH(aq) <> K₂SO₄(s) + 2H₂O(l)

Now that we have the reaction, we can go to the second part of the question.

To calculate the remaining concentration after neutralization, we need to calculate the moles of the reactants and determine which is the limiting reactant.

The moles of the reactants:

moles A = 0.42 * 0.15 = 0.063 moles

moles B = 0.210 * 0.1 = 0.021 moles

Now that we have the moles, let's calculate the limiting reactant:

H₂SO₄(aq) + 2KOH(aq) <> K₂SO₄(s) + 2H₂O(l)

If:

1 moles A > 2 moles B

0.063 A > X

X = 0.063 * 2 = 0.126 moles of B

However, we only have 0.021 moles of base, so, this is the limiting reactant.

Now that we know this, let's see the remaining moles of the acid, after the base reacts completely:

moles of A remaining = 0.063 - 0.021 = 0.042 moles

Finally to get the concentration, we have the volume of acid and the base together, so, the final volume is 0.25 L:

C = 0.042 / 0.25 = 0.168 M

This is the final concentration of the acid

58750.08068 is the answer i think