Question

1). The volume of a stock 6 M NaOH solution required to prepare 250 mL of a 1 M NaOH solution.

2). The mass of potassium hydrogen phthalate (KHP) required to react with 25 mL of a 1 M NaOH solution. The molecular weight of KHP is 204.23 g/mol.

3). The amount of water required to dissolve the quantity of KHP calculated in (b), given that the solubility of KHP is 10.2 grams per liter.

Answer 1

1) Dilution of NaOH:

Molarity of final solution=Molarity of stock*dilution factor=6M*(Vstock/250ml)

or1M=6M*(Vstock/250ml)

Vstock=41.7 ml

2) The rxn between NaOH and KHP:

NaOH(aq) +KHC8H4O4(aq)----->KNaC8H4O4(aq)+H2O(l)

NaOH and KHP reacts in 1:1 molar ratio, so mol of KHP required to react with NaOH=mol of NaOH

mol of NaOH required=0.025L*1M=0.025mol

mol of KHP required to react with NaOH=0.025 mol=mass of KHP/MW (molecular weight of KHP)=mass of KHP/(204.23 g/mol)

or, 0.025mol=KHP/(204.23 g/mol)

KHP=0.025mol*(204.23 g/mol)=5.106 g

3) solubility of KHP = 10.2 grams/L

volume of water required to dissolve=5.106g/(10.2 grams/L)=0.5L=500 ml