Question

In this problem, submit one part at a time, and use the given feedback to answer subsequent parts.

100. mL of aqueous 0.40 M CuSO4 is mixed with 100. mL of aqueous 0.20 M Na2S.

a. What kind of reaction could potentially take place?

-Bronsted-Lowry acid-base

-no reaction correct

-precipitation

-redox

b. Which of the following is a spectator ion in this mixture? The table at the bottom may be useful.(Select all that apply.)

- Na+

-SO42-

- S2-

-Cu2+

c. After the solutions are mixed, what will be the concentration of Na+?

-0.80 M

-0.10 M

-0.40 M

-0.20 M

d. After the solutions are mixed, what will be the concentration of SO42-?

-0.10 M

-0.80 M

-0.20 M

-0.40 M

e. Which reactant will limit the amount of product that can form?

-something else

-Cu2+

-S2-

f. This mixture will establish equilibrium in mere fractions of a second. After it does, what will be the concentration of Cu2+?

-0.20 M

-zero M

-0.40 M

-something much smaller than 0.10 M

-0.10 M

g. This mixture will establish equilibrium in mere fractions of a second. After it does, what will be the concentration of S2-? M

Answer 1

100. mL of aqueous 0.40 M CuSO4 is mixed with 100. mL of aqueous 0.20 M Na2S.

Note that Cu2+ and S-2 are present, they will form CuS(s) since CuS is not that soluble

a. What kind of reaction could potentially take place?

-precipitation

b. Which of the following is a spectator ion in this mixture? The table at the bottom may be useful.(Select all that apply.)

spectator ion are those which will NOT form any kind of precipitate, therefore, they are SO4-2 and Na2+ ions, since Na2SO4 is soluble

c. After the solutions are mixed, what will be the concentration of Na+?

[Na+] = M1*V1/(V1+V2) = 0.4*100/(100+100) = 0.2 M (halved)

d. After the solutions are mixed, what will be the concentration of SO42-?

[SO4-2] = M2*V2/(V1+V2) = 0.4*100/(100+100) = 0.2 M (halved)

e. Which reactant will limit the amount of product that can form?

mmol of CusO4 = 40

mmol of Na2S = 100*0.2 = 20

then,

ratio is 1;1 so S-2 is limiting reaciton

f. This mixture will establish equilibrium in mere fractions of a second. After it does, what will be the concentration of Cu2+?

[Cu2+] = 40 mmol - 20 mmol reacted = 20 mmol left

Vtotal = 100+100 = 200 mL

[Cu2+] = 20/200 = 0.1 M

g. This mixture will establish equilibrium in mere fractions of a second. After it does, what will be the concentration of S2-? M

the concentration will be mostly zero, since CuS <-> Cu2+ S-2 solbuiltiy is too low