Question

How many milliliters of 0.350 M HCl are needed to titrate each of the following solutions to the equivalence point?

(a) 49.4 mL of 0.455 M LiOH

__________mL

(b) 75.0 mL of 0.455 M NaOH

________mL

(c) 490.0 mL of a solution that contains 14.3 g of RbOH per liter

__________mL

Answer 1

a)

NaOH + HCL ---> NaCl + H20

we can see that

at equivalence point

moles of HCl = moles of LiOH added

we know that

moles = concentration x volume

so

Ma x Va = Mb x Vb

0.35 x Va = 0.455 x 49.4

Va = 64.22

so

64.22 ml of HCl are needed

b)

in this case also

Ma x Va = Mb x Vb

0.35 x Va = 0.455 x 75

Va = 97.5

so

97.5 ml of HCl is needed

c)

now

moles= mass / molar mass

so

moles of RbOH = 14.3 / 102.475

moles of RbOH = 0.13954623

now

concentration = moles / volume (L) = 0.13954623 / 1 = 0.13954623 M

now

Ma x Va = Mb x Vb

0.35 x Va = 0.13954623 x 490

Va = 195.36

so

195.36 ml of HCl is needed