Question

1. 20.00 mL of 0.510 M NaOH is titrated with 0.740 M H₂SO₄. ______ 1 mL of H₂SO₄ are needed to reach the end point.

2.  If 35.00 mL of 0.0200 M aqueous HCl is required to titrate 30.00 mL of an aqueous solution of NaOH to the equivalence point, the molarity of the NaOH solution is ________ 1 M.

Answer 1

Part A

first write the balanced equation

H2SO4 + 2NaOH -------> Na2SO4 + 2H2O

from the balanced equation it is clear that one mole of H2SO4 required 2 moles of NaOH

no of moles of NaOH = molarity x volume of NaOH in liters = 0.51 x 0.02 = 0.0102 moles

no of moles of H2SO4 required for 0.0102 moles of NaOH = 0.0102 /2 = 0.0051 moles

now we have moles of H2SO4 = 0.0051 and molarity of H2SO4 = 0.74 M

Volume of HCl = no of moles / Molarity = 0.0051 / 0.74 = 0.007L

= 0.007 x 1000 ml

= 7 ml of H2SO4

Part-2

first write the balanced equation

NaOH + HCl ------> NaCl + H2O

no o fmoles of HCl = molarity x volume in liters = 0.02M x 0.035L = 0.0007 moles

from the balanced equation one mole oh HCl required one mole NaOH

so no of moles of NaOH = 0.0007, volume of NaOH given = 30.0 mL = 0.03L

Molarity = moles/liters = 0.0007 mol / 0.03 L

= 0.023M