Question

how many milliliters of 0.15 M HCl are needed to titrate to the end point 100.0 mL of a solution that contains 2.35g of NaOH per liter?

Answer #1

Balanced chemical equation is:

NaOH + HCl ---> NaCl + H2O

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass(NaOH)= 2.35 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(2.35 g)/(40 g/mol)

= 5.875*10^-2 mol

According to balanced equation

mol of HCl reacted = (1/1)* moles of NaOH

= (1/1)*5.875*10^-2

= 5.875*10^-2 mol

This is number of moles of HCl

use:

M = number of mol / volume in L

0.15 = 5.875*10^-2/ volume in L

volume = 0.392 L

volume = 392 mL

Answer: 392 mL

How many milliliters of 0.170 M HCl are needed to titrate
each of the following solutions to the equivalence point?
(a) 42.6 mL of 0.0850 M RbOH
mL
(b) 24.6 mL of 0.102 M NaOH
mL
(c) 346.0 mL of a solution that contains 1.91 g of KOH per
liter
mL

How many milliliters of 0.350 M HCl are needed to titrate each
of the following solutions to the equivalence point?
(a) 49.4 mL of 0.455 M LiOH
__________mL
(b) 75.0 mL of 0.455 M NaOH
________mL
(c) 490.0 mL of a solution that contains 14.3 g of RbOH per
liter
__________mL

How many milliliters of 0.731 M HIO4 are needed to titrate each
of the following solutions to the equivalence point?
(a) 40.0 mL of 0.731 M RbOH
mL=
(b) 28.6 mL of 0.950 M KOH
mL =
(c) 798.0 mL of a solution that contains 8.76 g of NaOH per
liter
mL=

How many milliliters of 0.270 M HBr are needed to titrate each
of the following solutions to the equivalence point?
(a) 70.9 mL of 0.189 M RbOH mL
(b) 46.2 mL of 0.297 M CsOH mL
(c) 682.0 mL of a solution that contains 6.06 g of KOH per liter
mL

How many milliliters of 0.666 M HBr are needed to titrate
each of the following solutions to the equivalence point?
(a) 75.4 mL of 0.333 M LiOH
____________ mL
(b) 21.8 mL of 0.666 M KOH
_____________mL
(c) 102.0 mL of a solution that contains 34.1 g of RbOH per
liter
___________mL

How many milliliters of 8.00×10−2 M NaOH are required
to titrate each of the following solutions to the equivalence
point?
40.0 mL of 9.50×10−2 M HNO3
35.0 mL of 8.00×10−2 M HC2H3O2
55.0 mL of a solution that contains 1.85 g of HCl per liter

(a) How many milliliters of 0.165 M HCl are needed to
neutralize completely 35.0 mL of 0.101 M
Ba(OH)2 solution?
________ ml
(b) How many milliliters of 2.50 M
H2SO4 are needed to neutralize 50.0 g of
NaOH?
_________ mL
(c) If 56.8 mL of BaCl2 solution is needed to
precipitate all the sulfate in a 544 mg sample of
Na2SO4 (forming BaSO4), what is
the molarity of the solution?
_________M
(d) If 47.5 mL of 0.250 M HCl solution...

(a) How many milliliters of 0.165 M HCl are needed to neutralize
completely 35.0 mL of 0.101 M Ba(OH)2 solution? ______ml
(b) How many milliliters of 3.50 M H2SO4 are needed to
neutralize 75.0 g of NaOH?
_______mL
(c) If 55.8 mL of BaCl2 solution is needed to precipitate all
the sulfate in a 544 mg sample of Na2SO4 (forming BaSO4), what is
the molarity of the solution?
________M
(d) If 47.5 mL of 0.375 M HCl solution is needed...

Calculate how many milliliters of 0.100 M HCl solution is needed
to make a pH 4.34 solution with 2,2'-bipyridine, to 100.0 ml of
10.0 X 10-3 M of 2,2'-bipyridine.

You titrate 25.0 mL of 0.15 M HCl with 0.10 M sodium
hydroxide.
a. What is the pH after 20.0 mL NaOH have been added?
b. How many mL of NaOH are required to reach the equivalence
point?

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