In: Chemistry
100 mL of LiOH(aq) (0.500 M) is reacted with 250 mL of sulfuric acid solution (0.200 M). Consider the following statements;
i) This is an example of a strong acid-strong base neutralization reaction.
ii) LiOH(aq) is the limiting reagent and produces 0.05 moles of water
iii) The final concentration of Li+ (aq) in the mixture would be 0.071 M
a. Only statement i is correct.
b. Only statement ii is correct.
c. Only statement iii is correct.
d. Statements i and ii are correct.
e. Statements i, ii and iii are all correct.
Correct answer is : (d.) Statements i and ii are correct.
Explanation
Lithium hydroxide LiOH is a strong base and sulfuric acid H2SO4 is a strong acid. Hence this is an example of a strong acid-strong base neutralization reaction.
The balanced reaction equation is : 2 LiOH(aq) + H2SO4(aq) Li2SO4(aq) + 2 H2O(l)
moles LiOH = (concentration LiOH) * (volume LiOH in Liter)
moles LiOH = (0.500 M) * (0.100 L)
moles LiOH = 0.050 mol
moles H2SO4 = (concentration H2SO4) * (volume H2SO4 in Liter)
moles H2SO4 = (0.200 M) * (0.250 L)
moles H2SO4 = 0.050 mol
LiOH is the limiting reactant because from the balanced equation one mole of H2SO4 requires 2 moles of LiOH. By this ratio, 0.050 moles of H2SO4 will require (2 * 0.050 moles) = 0.100 moles LiOH but we have only 0.050 moles LiOH.