Question

How many milliliters of 0.170 M HCl are needed to titrate each of the following solutions to the equivalence point?




(a) 42.6 mL of 0.0850 M RbOH

mL



(b) 24.6 mL of 0.102 M NaOH

mL



(c) 346.0 mL of a solution that contains 1.91 g of KOH per liter

mL

Answer 1

V1 x N1 = V2 x N2

(a) 42.6 mL of 0.0850 M RbOH

V1 = 42.6 ml N1 = 0.085

N2 = 0.170 V2 = ?

V2 = 42.6 x 0.085 / 0.17 = 21.3 ml

21.3 ml of 0.17 M HCl is need to equivalence point

(b) 24.6 mL of 0.102 M NaOH

V1 = 24.6 ml N1 = 0.102

N2 = 0.170 V2 = ?

V2 = 24.6 x 0.102 / 0.17 = 14.76 ml

14.76 ml of 0.17 M HCl is need to equivalence point

346.0 mL of a solution that contains 1.91 g of KOH per liter

Concentration of KOH = 1.91 x 1000 /56.12 x 346 = 0.09836 M

V1 = 346 N1 = 0.09836

V2 = ? N2 = 0.17 M

V2 = 346 x 0.09836 / 0.17 = 200.12 ml

200.12 ml of 0.17 M HCl is need to equivalence point