how many milliliters of 0.250M KOH will react with 15.0mL of 0.305 M H2SO4?
To determine the number of milliliters of 0.250 M KOH that will react with 15.0 mL of 0.305 M H2SO4, we need to use stoichiometry and the balanced equation for the reaction between KOH and H2SO4.
First, let's write the balanced equation:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of H2SO4.
Using the molarity of H2SO4 and its volume, we can find the number of moles present:
moles of H2SO4 = Molarity * volume
= 0.305 M * 15.0 mL
= 4.575 mmol
Since the molar ratio between KOH and H2SO4 is 2:1, we can determine the number of moles of KOH needed:
moles of KOH = 1/2 * moles of H2SO4
= 1/2 * 4.575 mmol
= 2.2875 mmol
Now, we can calculate the volume of 0.250 M KOH needed to supply this amount of KOH:
volume of KOH = moles of KOH / Molarity
= 2.2875 mmol / 0.250 M
= 9.15 mL
Therefore, 9.15 milliliters of 0.250 M KOH will react with 15.0 milliliters of 0.305 M H2SO4.