Question

How many milliliters of 0.640 M HCl are needed to titrate each of the following solutions to the equivalence point?

(a) 30.2 mL of 0.640 M LiOH

_________ mL

(b) 60.8 mL of 0.768 M NaOH

________ mL

(c) 464.0 mL of a solution that contains 6.56 g of RbOH per liter

________ mL

Answer 1

number of mols of HCl needed = number of mols of base present

a)
number of mols of LiOH = volume of LiOH * molarity
= 30.2 ml * 0.640 M = 19.328 milli mols

so number of mols of HCl needed =19.328 milli mols

volume of HCl needed = number of mols of HCl/molarity of HCl
= 19.328 milli mols/0.640 M
=30.2 ml
round to 3 sig.fig
volume of HCl needed = 30.2 ml
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b)
number of mols of NaOH = volume of NaOH * molarity
=60.8 ml * 0.768 M =46.6944 milli mols

so number of mols of HCl needed =46.6944 milli mols

volume of HCl needed = number of mols of HCl/molarity of HCl
= 46.6944 milli mols/0.640 M
=72.96ml
round to 3 sig.fig
volume of HCl needed = 73.0 ml
*****************
c)
molar mass of RbOH =
mass of RbOH present = 6.56 g
number of mols of RbOH = mass/molar mass = 6.56 g/102.48 g/mol
= 0.06401249 mols
molarity of RbOH = mols of RbOH/volume of solution
=0.06401249 mols/1 L
= 0.06401249 M

mols of RbOH present in 464 ml solution = 0.06401249 M * 464 ml
= 29.70179 mols

volume of HCl needed =29.70179 mols/0.640 M = 46.4 ml

volume of HCl = 46.4 ml

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