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If 10.0mL. of phosphoric acid solution required 23.5mL of 0.200M sodium hydroxid

ID: 983450 • Letter: I

Question

If 10.0mL. of phosphoric acid solution required 23.5mL of 0.200M sodium hydroxide solution for neutralization to a phenolphthalein end pointt, calculate the molarity of the phosphoric acid. H3PO4(aq) + 3NaOH(aq) rightarrow Na3PO4(aq) + 3H20(1) a. 0.157 M b. 0.470 M c. 1.41 M f. 1.17 M e. none of these 11. What mass of NaOH is needed to precipitate the Cd^- ions from 35.0 mL of 0.500 M Cd(NO_3)_2 solutic Cd(NO3)_2 (aq) + 2 NaOh_(aq) rightarrow 2 NaNO3_(aq) + Cd(OH)_2 (s) a. 0.0175 g b. 1.40 g c. 0.0350 g d. 0.700 g e. none of these

Explanation / Answer

10) consider the given reaction

H3P04 + 3 NaOH ---> Na3P04 + 3 H20

we can see that

moles of NaOH = 3 x moles of H3P04

also

moles = molairty x volume (L)

so

M x V of NaOH = 3 x M x V of H3P04

so

0.2 x 23.5 = 3 x M x 10

M = 0.157

so

a) 0.157 M

11) we can see that

moles of NaoH = 2 x moles of Cd(N03)2

so

moles of NaOH = 2 x M x V of Cd(N03)2

moles of NaoH = 2 x 0.5 x 35 x 10-3

moles of NaOH = 35 x 10-3

now

mass = moles x molar mass

so

mass of NaOH = 35 x 10-3 x 40

mass of NaOH = 1.4 g

so

b) 1.40 g


12)

moles of Mg(OH)2 = 2.87 / 58.3197

moles of Mg(OH)2 = 0.0492

now

the reaction is =

Mg(OH)2 + 2HCl ---> MgCl2 + 2H20

so

moles of HCl = 2 x moles of Mg(OH)2

moles of HCl = 2 x 0.0492

moles of HCl = 0.0984

now

volume = moles / molarity

so

volume of HCl = 0.0984 / 0.128

volume of HCl = 0.769 L

volume of HCl = 769 ml

so

d) 769 ml


13) the reaction is

HCl04 + NaOH ---> NaCl04 + H20

so

moles of HCl04 = moles of NaoH

M x V of HCl04 = M x V of NaoH

so

0.115 x V = 0.0875 x 50

V = 38.04 ml

so

C) 38 ml


14)

moles of CaCl2 = 785 x 10-3 / 111 = 7.072 x 10-3

now

moles of Cl- = 2 x 7.072 = 14.144 x 10-3

now

the reaction is

Ag+ + Cl- --> AgCl

so

moles of Ag+ = moles of Cl-

moles of Ag+ = 14.144 x 10-3

now

moles of AgN03 = moles of Ag+ = 14.144 x 10-3

so

molarity = moles / volume (L)

so

molarity = 14.144 x 10-3 / 25.8 x 10-3

molarity = 0.548

so

a) 0.548 M


15)

the reaction is

2 Fe + 3Br2 ---> 2 FeBr3


Fe ---> Fe+2 so Fe is oxidized , so reducing agent

Br2 --> Br-   so Br is reduced , so oxidizing agent


so

the answer is option B

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