how many milliliters of a 2.61 m h2so4 solution are needed to neutralize 71.50 ml of a 0.855 m koh solution
To determine the volume of the H2SO4 solution needed to neutralize the KOH solution, we need to use the concept of stoichiometry, which is the study of the quantitative relationship between the reactants and products in a chemical reaction.
First, let's write the balanced chemical equation for the reaction between H2SO4 and KOH:
H2SO4 + 2KOH -> K2SO4 + 2H2O
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH, producing 1 mole of K2SO4 and 2 moles of H2O.
Next, we need to calculate the number of moles of KOH in the given 71.50 mL of 0.855 M KOH solution:
moles of KOH = volume of KOH solution (in liters) × molarity of KOH
Convert the volume of KOH solution from milliliters to liters:
71.50 mL = 71.50 mL / 1000 mL/L = 0.07150 L
Now substitute the values into the formula:
moles of KOH = 0.07150 L × 0.855 M = 0.06114375 moles
Since the stoichiometry of the balanced equation shows a 1:2 mole ratio between H2SO4 and KOH, we can determine the number of moles of H2SO4 needed to neutralize the given amount of KOH:
moles of H2SO4 = 0.06114375 moles × (1 mole H2SO4 / 2 moles KOH) = 0.030571875 moles
Finally, we can calculate the volume of the 2.61 M H2SO4 solution needed to provide the equivalent number of moles:
volume of H2SO4 solution = moles of H2SO4 / molarity of H2SO4
Substitute the values into the formula:
volume of H2SO4 solution = 0.030571875 moles / 2.61 M = 0.011717644 L
Convert the volume of H2SO4 solution from liters to milliliters:
volume of H2SO4 solution = 0.011717644 L × 1000 mL/L = 11.7176 mL
Therefore, approximately 11.7176 milliliters of the 2.61 M H2SO4 solution are needed to neutralize 71.50 mL of the 0.855 M KOH solution.