How many mL of 1.20 M H2SO4 are needed to neutralize 18.75 mL of 0.843 M LiOH?
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6.59
To determine the volume of H2SO4 needed to neutralize LiOH, we first need to find the balanced chemical equation for the neutralization reaction between H2SO4 and LiOH.
The balanced chemical equation for the neutralization reaction is as follows:
H2SO4 + 2LiOH -> Li2SO4 + 2H2O
From the balanced equation, we can see that one mole of H2SO4 reacts with two moles of LiOH to produce one mole of Li2SO4 and two moles of water (H2O).
Step 1: Calculate the number of moles of LiOH:
Using the given concentration and volume, we can calculate the number of moles of LiOH using the formula:
moles = concentration x volume
moles of LiOH = 0.843 M x 0.01875 L = 0.0158 moles
Step 2: Determine the mole ratio between H2SO4 and LiOH:
From the balanced equation, we know that for every 2 moles of LiOH, we need 1 mole of H2SO4. Therefore, the mole ratio between H2SO4 and LiOH is 1:2.
Step 3: Calculate the number of moles of H2SO4 needed:
Since the mole ratio between H2SO4 and LiOH is 1:2, the number of moles of H2SO4 needed is half the number of moles of LiOH.
moles of H2SO4 = 0.0158 moles / 2 = 0.0079 moles
Step 4: Calculate the volume of 1.20 M H2SO4 needed:
Using the concentration and the number of moles of H2SO4, we can calculate the volume of 1.20 M H2SO4 using the formula:
volume = moles / concentration
volume of H2SO4 = 0.0079 moles / 1.20 M = 0.0066 L or 6.6 mL
Therefore, 6.6 mL of 1.20 M H2SO4 are needed to neutralize 18.75 mL of 0.843 M LiOH.