Question

Part A: What volume of 0.105 M HClO4 solution is needed to neutralize 50.00 mL of 8.75x10^-2 M NaOh? in mL

Part B: What volume of 0.130 M HCl is needed to neutralize 2.60g of Mg(OH)2?

Part C: If 27.0 mL of AgNO3 is needed to precipitate all the Cl- ions in a 0.785mg sample of KCl (forming AgCl) what is the molarity of the AgNO3 solution?

Part D: If 45.0 mL of 0.110 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Answer 1

a) The volume of acid needed is calculated:

Va = Cb * Vb / Ca = 8.75x10 ^ -2 M * 50 mL / 0.105 M =

b) The volume of acid is calculated:

Va = 2.6 g Mg (OH) 2 * (1 mol Mg (OH) 2 / 58.3 g) * (2 mol HCl / 1 mol Mg (OH) 2) * (1000 mL HCl / 0.13 mol) = 686 mL

c) The moles of KCl are calculated:

n KCl = g / MM = 0.000785 g / 74.55 = 1.1x10 ^ -5 mol

The molarity is calculated:

M AgNO3 = n / V = ​​1.1x10 ^ -5 / 0.027 L = 4x10 ^ -4 M

d) The mass of KOH is calculated:

m KOH = M * V * MM = 0.11 * 0.045 * 56.1 = 0.278 g

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