Question

In: Chemistry

100 mL of LiOH(aq) (0.500 M) is reacted with 250 mL of sulfuric acid solution (0.200...

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.

Solutions

Expert Solution

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.


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