Question

a)What volume of 0.122 M HCl is needed to neutralize 2.61 g of Mg(OH)2?

b)If 25.4 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.755-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution?

C)If 45.1 mL of 0.102 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Answer 1

One mole of Mg(OH)2 requires 2 mole of HCl for neutralization

Number of moles of Mg(OH)2 = mass/molar mass = 2.61/58.3 = 0.044768 moles

Moles of HCl = 2 * 0.044768 = 0.089536 moles

V/1000 * 0.122 = 0.089536

V = 733.90 mL

b) Molar mass of KCl = 74.55 gm/mol

number of moles of KCl = 0.755 * 10^(-3)/74.55 = 1.0127 * 10^(-5) moles

25.4/1000 * M = 1.0127 * 10^(-5)

M = 3.9871 * 10^(-4) M

c) KOH + HCl ---> KCl + H2O

Number of moles of HCl = 45.1/1000 * 0.102 = 0.0046002 moles

Molar mass of KOH = 39.09+16+1 = 56.09 gms/mole

Mass of KOH required = 0.0046002 moles * 56.09 gms/mole = 0.2580 gms