It takes 75 mL of 0.124 M KOH to neutralize 28.5 mL of H2SO4 solution. What is the concentration of the H2SO4 solution?
Answer in moles/liter.
To find the concentration of the H2SO4 solution, we can use the concept of stoichiometry and the balanced chemical equation:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
From the equation, we can see that it takes 2 moles of KOH to react with 1 mole of H2SO4.
First, let's calculate the number of moles of KOH used:
moles KOH = volume of KOH solution (in liters) x concentration of KOH (in moles/liter)
= 75 mL x 0.124 mol/L
= 9.3 mmol (millimoles)
Since the ratio between H2SO4 and KOH is 1:2, the number of moles of H2SO4 in the reaction is half that of KOH:
moles H2SO4 = 9.3 mmol / 2
= 4.65 mmol
Now, let's calculate the concentration of the H2SO4 solution:
concentration of H2SO4 = moles H2SO4 / volume of H2SO4 solution (in liters)
= 4.65 mmol / 0.0285 L
= 163.16 mmol/L
Finally, let's convert the concentration to moles/liter:
concentration of H2SO4 = 163.16 mmol/L
= 0.163 mol/L
Therefore, the concentration of the H2SO4 solution is 0.163 mol/L.
To find the concentration of the H2SO4 solution, we need to use the concept of stoichiometry. The balanced chemical equation for the reaction between KOH and H2SO4 is:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
From the equation, we can see that 2 moles of KOH react with 1 mole of H2SO4.
Given that it takes 75 mL of 0.124 M KOH to neutralize 28.5 mL of H2SO4 solution, we can set up the following proportion:
(0.124 M KOH) / (2 moles KOH) = (x M H2SO4) / (1 mole H2SO4)
Cross-multiplying, we get:
(0.124 M KOH) * (1 mole H2SO4) = (x M H2SO4) * (2 moles KOH)
Simplifying, we find:
0.124 * 1 = 2 * x
x = 0.124 / 2
x = 0.062 moles/liter
Therefore, the concentration of the H2SO4 solution is 0.062 moles/liter.