Top topic this week
Back to FAQs
Related FAQs
Frequently Asked Question
Questions asked by users might not always be phrased in the clearest way.
A 150 ml sample of H2SO4 solution from an automobile storage battery is titrated with 0.5 m NaOHrnsolution to a phenolphthalein end point, requiring 24 ml. What is the molarity of H2SO4 solution?rn[2 M]rn(A)rn23.2 Mrn(B)rn11.76 Mrn(C)rn0.04 Mrn(D)rn6 m
Accepted Answer
Balanced chemical equation:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
Moles of NaOH used:
Moles of NaOH = Molarity x Volume(in L)
Moles of NaOH = 0.5 M x 0.024 L
Moles of NaOH = 0.012 moles
Moles of H2SO4:
From the balanced chemical equation, 2 moles of NaOH react with 1 mole of H2SO4.
Moles of H2SO4 = 0.012 moles / 2
Moles of H2SO4 = 0.006 moles
Molarity of H2SO4 solution:
Molarity = Moles of solute / Volume of solution (in L)
Molarity of H2SO4 = 0.006 moles / 0.15 L
Molarity of H2SO4 = 0.04 M
Answer: (C) 0.04 M
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
Moles of NaOH used:
Moles of NaOH = Molarity x Volume(in L)
Moles of NaOH = 0.5 M x 0.024 L
Moles of NaOH = 0.012 moles
Moles of H2SO4:
From the balanced chemical equation, 2 moles of NaOH react with 1 mole of H2SO4.
Moles of H2SO4 = 0.012 moles / 2
Moles of H2SO4 = 0.006 moles
Molarity of H2SO4 solution:
Molarity = Moles of solute / Volume of solution (in L)
Molarity of H2SO4 = 0.006 moles / 0.15 L
Molarity of H2SO4 = 0.04 M
Answer: (C) 0.04 M
Articles you might like