If 22.15 milliliters of 0.100 molarity sulfuric acid is required to neutralize 10.0 milliliters of lithium hydroxide solution, what is the molar concentration of the base?
2LiOH + H2SO4 ==> 2H2O + Li2SO4
mols H2SO4 = M x L = ?
Convert mols H2SO4 to mols LiOH (use the coefficients in the balanced equation to do this).
Then M LiOH = mols LiOH/L LiOH.
To find the molar concentration of the base (lithium hydroxide solution), you need to use the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and lithium hydroxide (LiOH). The balanced equation is as follows:
H2SO4 + 2LiOH → Li2SO4 + 2H2O
From the balanced equation, you can see that one mole of sulfuric acid reacts with two moles of lithium hydroxide.
First, convert the volume of sulfuric acid (22.15 milliliters) to moles using the given molar concentration (0.100 M).
moles of H2SO4 = volume (liters) x molarity
moles of H2SO4 = 0.02215 liters x 0.100 moles/liter
moles of H2SO4 = 0.002215 moles
From the balanced equation, you can see that the mole ratio between sulfuric acid and lithium hydroxide is 1:2. Therefore, the number of moles of lithium hydroxide used in the neutralization reaction is twice the number of moles of sulfuric acid.
moles of LiOH = 2 x moles of H2SO4
moles of LiOH = 2 x 0.002215 moles
moles of LiOH = 0.00443 moles
Next, calculate the molar concentration of the base (LiOH) by dividing the moles of LiOH by the volume of the lithium hydroxide solution (10.0 milliliters).
molarity of LiOH = moles of LiOH / volume (liters)
molarity of LiOH = 0.00443 moles / 0.010 liters
molarity of LiOH = 0.443 M
Therefore, the molar concentration of the lithium hydroxide solution is 0.443 M.