In this problem, submit one pa
In this problem, submit one part at a time, and use the givenfeedback to answer subsequent parts.
100. mL of aqueous 0.40 M CuSO4 is mixed with 100. mL ofaqueous 0.20 M Na2S.
a. What kind of reaction could potentially takeplace?
-Bronsted-Lowry acid-base
-no reaction correct
-precipitation
-redox
b. Which of the following is a spectator ion in thismixture? The table at the bottom may be useful.(Select all thatapply.)
– Na+
-SO42-
– S2-
-Cu2+
c. After the solutions are mixed, what will be theconcentration of Na+?
-0.80 M
-0.10 M
-0.40 M
-0.20 M
d. After the solutions are mixed, what will be theconcentration of SO42-?
-0.10 M
-0.80 M
-0.20 M
-0.40 M
e. Which reactant will limit the amount of product thatcan form?
-something else
-Cu2+
-S2-
f. This mixture will establish equilibrium in merefractions of a second. After it does, what will be theconcentration 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 merefractions of a second. After it does, what will be theconcentration of S2-? M
Answer:
100. mL of aqueous 0.40 M CuSO4 is mixed with 100. mL ofaqueous 0.20 M Na2S.
Note that Cu2+ and S-2 are present, they will form CuS(s) sinceCuS is not that soluble
a. What kind of reaction could potentially takeplace?
-precipitation
b. Which of the following is a spectator ion in thismixture? The table at the bottom may be useful.(Select all thatapply.)
spectator ion are those which will NOT form any kind ofprecipitate, therefore, they are SO4-2 and Na2+ ions, since Na2SO4is soluble
c. After the solutions are mixed, what will be theconcentration 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 theconcentration 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 thatcan 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 merefractions of a second. After it does, what will be theconcentration 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 merefractions of a second. After it does, what will be theconcentration of S2-? M
the concentration will be mostly zero, since CuS <-> Cu2+S-2 solbuiltiy is too low
"Our Prices Start at $11.99. As Our First Client, Use Coupon Code GET15 to claim 15% Discount This Month!!"
