Question

Two 25.0-mL aqueous solutions, labeled A and B, contain the ions indicated:

Answer 1

The additional volume of HCl which must be added to reach to the equivalence point is 8.33 mL

The moles of HCl which is required to reach the equivalence point can be calculated in the way as follows.

Moles of HCl can be calculated as

Moles of HCl = 0.004 moles of Ca (OH) 2 × 2 moles of HCl / 1 moles of Ca (OH) 2

= 0.008 moles of HCl

The volume of HCl which is required to reach the equivalence point can be calculated in the way given as follows.

Volume of HCl required= 0.008 moles of HCl × 1 L / 0.24 moles of HCl × 1 ml / 10 -³ L

= 33.33 ml

The additional volume of HCl calculated as

Additional volume = required volume – actual volume

= 33.33 mL – 25 mL

= 8 . 33 mL

Thus, we calculated that the additional volume of HCl which must be added to reach to the equivalence point is 8.33 mL.

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