Question

In: Chemistry

A solution is made of 1.1 x 10-3M in Zn(NO3)2 and 0.150 M in NH3. After...

A solution is made of 1.1 x 10-3M in Zn(NO3)2 and 0.150 M in NH3.

After the solution reaches equilibrium, what concentration of Zn2+(aq) remains?

Kf (Zn(NH3)42+) = 2.8 x 109

Solutions

Expert Solution

8.74×10-10M

Explanation

Consider the complete formation of Zn(NH3)42+

Zn2+(aq) + 4NH3(aq) ------> Zn(NH3)42+(aq)

stoichiometrically , 1mole of Zn2+ reacts with 4moles of NH3 to give 1mole of Zn(NH3)42+

0.0011M of Zn2+ reacts with 0.0044M of NH3 to give 0.0011M of Zn(NH3)42+

After reaction

Remaining concentration of NH3 = 0.1500M - 0.0044M = 0.1456M

Now, consider the dissociation equillibrium of Zn(NH3)42+

Zn(NH3​​​​​​)42+(aq) <---------> Zn2+(aq) + 4NH3(aq)

Kd = [Zn2+][NH3]4/[Zn(NH3)42+]

Kd = 1/Kf

Kd = 1/ 2.8 ×109 = 3.57×10-10

Initial concentrations

[Zn(NH3)42+] = 0.0011

[Zn2+] = 0

[NH3] = 0.1456

Change in concentration

[Zn(NH3)42+] = -x

[Zn2+] = +x

[NH3] = +4x

Equillibrium concentration

[Zn(NH3)42+] = 0.0011 - x

[Zn2+] = x

[NH3] =0.1456 + 4x

so,

x(0.1456 + 4x)4/(0.0011 - x) = 3.57 ×10-10

we can assume 0.1456 + 4x = 0.1456 and 0.0011-x = 0.0011

x(0.1456)4/(0.0011) = 3.57 ×10-10

x0.4086 = 3.57 ×10-10

x = 8.74 ×10-10

Therefore,

at equillibrium ,

[Zn2+] = 8.74 ×10-10M

  

queen_honey_blossom answered 1 year ago
Next > < Previous

Related Solutions

A solution is made that is 1.1×10−3 M in Zn(NO3)2 and 0.140 M in NH3. Part...
A solution is made that is 1.1×10−3 M in Zn(NO3)2 and 0.140 M in NH3. Part A After the solution reaches equilibrium, what concentration of Zn2+(aq) remains? Express your answer using two significant figures.
A solution is made by mixing 0.150 M NH4Cl and 0.100 M NH3. Ka for NH4+...
A solution is made by mixing 0.150 M NH4Cl and 0.100 M NH3. Ka for NH4+ is 5.61x10-10 (please show math) a. What type of solution you will have when the two compounds are mixed? b. Identify the acid and the base (write the formula and the ID next to it)    c. Will Cl- affect the pH of the solution? Why?   d. What is your prediction for the calculated pH compared to the pKa of the sample? Why?   e. What...
1.A solution is made by mixing 0.150 M NH4Cl and 0.100 M NH3. Ka for NH4+...
1.A solution is made by mixing 0.150 M NH4Cl and 0.100 M NH3. Ka for NH4+ is 5.61x10-10 a.What type of solution you will have when the two compounds are mixed? b.Identify the acid and the base (write the formula and the ID next to it) c.Will Cl- affect the pH of the solution? Why? d.What is your prediction for the calculated pH compared to the pKa of the sample? Why? e.What is the pH of the solution? Was your...
A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added...
A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added until the pH of the solution is 8.50. Which of the following is true? Pb(OH)2, Ksp= 1.4 x10^-20 Mn(OH)2, Ksp= 2.0 x 10^-13 Zn(OH)2, Ksp= 2.1 x10^-16 A) No precipitate will form. B) Only Pb(OH)2 will precipitate. C) Only Mn(OH)2 wil precipitate. D) Only Zn(OH)2 and Pb(OH)2 will preipitate. E) All three hydroxide will preipitate.
Equal volumes of a 0.020 M Zn^2+ solution and a 2.0 M NH3 solution are mixed....
Equal volumes of a 0.020 M Zn^2+ solution and a 2.0 M NH3 solution are mixed. Kf for [Zn(NH3)4]^2+ is 4.1 × 10^8. If enough sodium oxalate is added to make the solution 0.10 M in oxalate, will ZnC2O4 precipitate? What is Q? Ksp ZnC2O4 = 2.7 × 10^-8 Answer: no, Q = 2.9 × 10^-12
The complex ion Cu(NH3)42 is formed in a solution made of 0.0400 M Cu(NO3)2 and 0.400...
The complex ion Cu(NH3)42 is formed in a solution made of 0.0400 M Cu(NO3)2 and 0.400 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42 at equilibrium? The formation constant*, Kf, of Cu(NH3)42 is 1.70 × 1013.
The complex ion Cu(NH3)42 is formed in a solution made of 0.0200 M Cu(NO3)2 and 0.400...
The complex ion Cu(NH3)42 is formed in a solution made of 0.0200 M Cu(NO3)2 and 0.400 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42 at equilibrium? The formation constant*, Kf, of Cu(NH3)42 is 1.70 × 1013.
The complex ion Cu(NH3)42 is formed in a solution made of 0.0400 M Cu(NO3)2 and 0.500...
The complex ion Cu(NH3)42 is formed in a solution made of 0.0400 M Cu(NO3)2 and 0.500 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42 at equilibrium? The formation constant*, Kf, of Cu(NH3)42 is 1.70 × 1013.
If a solution contains 7.5 X 10-8M Cu+2 and 1.0 X 10-7 M Zn+2 will either...
If a solution contains 7.5 X 10-8M Cu+2 and 1.0 X 10-7 M Zn+2 will either metal precipitate as a hydroxide salt if the pH is 8.5? At what concentration of OH- will each metal precipitate?
Consider the following reactions: Zn^2+ + 4H3 <----> Zn(NH3)4^2+ beta 4 = 5.01 x 10^8 Zn^2+...
Consider the following reactions: Zn^2+ + 4H3 <----> Zn(NH3)4^2+ beta 4 = 5.01 x 10^8 Zn^2+ + 2e^- <-----> Zn(s) E degree = -0.762 V Assuming there is negligible current and nearly all Zn^2+ is in the form Zn(NH3)4^2+. what cathode potential (vs. S.H.E.) would be required to reduce 99.99% of the Zn^2+ from a solution containing 0.20 M Zn^2+ in 1.4 M ammonia? (Assume T= 298 K) ________V
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT