Question

A 30.00 mL sample of an unknown HClO4 solution requires titration with 22.55 mL of 0.2096 M Sr(OH)2 to reach the equivalence point.

Answer 1

The balance chemcial reaction is as followS:

Sr(OH)2 (aq) + 2 HClO4 (aq) ----> Sr(ClO4)2 (aq) + 2 H2O (l)
Neutralization of Base and Acid ----> to yield Salt and Water .

Now calcualte the moles of Sr(OH)2 as follows:

Here;

0.2096 M Sr(OH)2 = 0.2096 mol /L Sr(OH)2 = 0.2096 moles / 1000 ml Sr(OH)2

Volume = 22.55 ml

Number of moles = volume * molarity

= 0.2096 moles / 1000 ml Sr(OH)2 * 22.55 ml

= 4.7*10^-3 mol

Now calcualte the mole of HClO4 which is used:

4.7*10^-3 mol Sr(OH)2 * 2 mol HClO4 / 1 mol Sr(OH)2

= 9.4 *10^-3 HClO4

Here volume of HClO4 = 30 ml or 0.030 L

Molarity of HClO4 = moles / volume

= 9.4 *10^-3 HClO4 /0.030 L

= 0.31 M