(A) How many milliliters of 0.135 M HCl are needed to completely neutralize 47.0

Question

(A) How many milliliters of 0.135 M HCl are needed to completely neutralize 47.0mL of 0.110 M Ba(OH)2 solution?

V = ??mL

Part B

How many milliliters of 0.123 M H2SO4 are needed to neutralize 0.240g of NaOH?

?? mL

Part C

If 57.0mL of BaCl2 solution is needed to precipitate all the sulfate ion in a 746mg sample of Na2SO4, what is the molarity of the solution?

Part D

If 42.5mL of 0.200 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution?

?? mL

Explanation / Answer

(a)

N1V1=N2V2

V2 = 2*0.11*47/0.135 = 76.593 ml

b)

2*0.123*x=0.240/40

x=0.0244 liters = 24.4 ml

c)

746/142 = x*57 so x = 0.0921 ml

d)

42.5*0.2 = 2*x/74

x = 314.5 mg = 0.3145 grams

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