100ml of 1.0 m KOH solution will neutralize 100 ml of 1.0 M H2SO4 solution
Thank you!
no, it won't
H2SO4 is diprotic (two H+ per molecule)
You're welcome! Let me explain the concept behind the given statement.
To determine the neutralization reaction, we need to understand the concept of molarity (M) and the balanced chemical equation between the two reactants.
Molarity (M) is a measure of the concentration of a solution. It is defined as the number of moles of solute per liter of solution. In this case, a 1.0 M KOH solution means that it contains 1.0 mole of KOH per liter of solution.
The balanced chemical equation for the neutralization reaction between KOH (potassium hydroxide) and H2SO4 (sulfuric acid) is:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
According to this equation, we can see that 2 moles of KOH reacts with 1 mole of H2SO4 to produce 1 mole of K2SO4 and 2 moles of water.
Since the molarity of both the KOH and H2SO4 solutions is 1.0 M, that means they both contain 1.0 mole of KOH and H2SO4 per liter of solution.
Given that 100 ml (0.1 L) of each solution is used, we can calculate the number of moles of KOH and H2SO4 present in the reaction.
Moles of KOH = Molarity × Volume = 1.0 M × 0.1 L = 0.1 moles
Moles of H2SO4 = Molarity × Volume = 1.0 M × 0.1 L = 0.1 moles
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of H2SO4. Therefore, if we have an equal number of moles of KOH and H2SO4, they will react in a 1:1 ratio.
Since both solutions contain 0.1 moles, the amount of KOH will be equal to the amount of H2SO4. Therefore, 100 ml of 1.0 M KOH solution will neutralize 100 ml of 1.0 M H2SO4 solution.
I hope this explains how to arrive at the answer! If you have any further questions, feel free to ask.