How many milliliters of 0.250 M KOH will react with 15.0 mL of 0.350 M H2SO4?
First, let us write down the chemical equation for this reaction.
`2 KOH + H_2SO_4 -> K_2SO_4 + 2 H_2O`
From the balanced chemical equation, we can see that 2 moles of KOH is needed to react 1 mole of H2SO4.
First, we need to get the moles of H2SO4:
`Mol es H_2SO_4 = (15.0)/(1000) L * 0.350M`
`Mol es H_2SO_4 = 0.00525 mol es H_2SO_4`
From the balanced equation:
`0.00525 mol es H_2SO_4 * (2 mol es KOH)/(1 mol e H_2SO_4)`
= 0.0105 moles KOH
Finally, to get the volume, we need the expression for molarity:
`Molarity = (mol es solute)/(Volume of Solution(L))`
`Volume of solution = (Mol es KOH)/(Molarity)`
`Volume of solution = (0.0105)/(0.250)`
Volume of solution = 0.042 L = 42 mL **
** 1000mL = 1L