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In the neutralization reaction

H₂SO₄+2KOH ⇋ 2H₂O+ K₂SO₄

If there is 909 mL of 0.210 M H₂SO₄ reacting with 427 mL of a KOH solution with unknown concentration, what is the concentration of KOH?

User Jeff Sharp
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Final answer:

The concentration of the KOH solution reacting with H2SO4 is found using stoichiometry of the balanced chemical reaction.

Step-by-step explanation:

The student is asking how to determine the concentration of a KOH solution when reacted with a given amount of H&sub2;SO&sub4;. We can use the stoichiometry of the balanced neutralization reaction, which tells us that it takes two moles of KOH to neutralize one mole of H&sub2;SO&sub4;. This reaction is represented as:

H&sub2;SO&sub4; + 2KOH → 2H&sub2;O + K&sub2;SO&sub4;

First, we calculate the moles of H&sub2;SO&sub4;:

Moles H&sub2;SO&sub4; = 0.909 L × 0.210 M = 0.19089 moles

Using the stoichiometry from the balanced equation, we need twice as many moles of KOH to react completely with H&sub2;SO&sub4;, so we need:

Moles KOH required = 2 × moles H&sub2;SO&sub4; = 2 × 0.19089 mol = 0.38178 moles

The volume of KOH has been given as 0.427 L. The concentration of KOH can be found using the formula:

Concentration KOH = moles KOH / volume KOH = 0.38178 mol / 0.427 L = 0.89345 M

Therefore, the concentration of KOH is 0.893 M.

By calculating the moles of H2SO4 and applying the stoichiometric ratio, we determine that the concentration of KOH is 0.893 M.

User Rosangela
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