I need some help understanding this.

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

2 years ago

Given the equilibrium reaction: 2 A (aq) + 3 B (aq) <— —> 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M,

Alex787 [66]

Given the equilibrium reaction: 2 A (aq) + 3 B (aq) <— —> 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M, [B] = 0.30 M, [C] = 0.10 M and [D] = 0.50 M. The Kc value will be:
a. 1.9 c. 2.4 b. 0.15 d. 0.51

Answer . A

5 0

3 years ago

Please help me with this question! Am I right?

muminat

You are right!! Good job!!!

5 0

2 years ago

Group___are the most stable elements

GarryVolchara [31]

Answer:

Group VIIIA in which the noble/inert gases are found

7 0

2 years ago

HELP ME I AM DUMB!!!!!!!!!!!!!!!!!!

densk [106]

Answer:

2H⁺(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2H₂O(ℓ)

Explanation:

You aren't dumb. You just need more time to learn the concepts.

There are three steps you must follow. You must write the:

Molecular equation
Ionic equation
Net ionic equation

1. Molecular equation

2HBr + Sr(OH)₂ ⟶ SrBr₂ + 2H₂O

To predict the states of the substances, we must remember some solubility rules:

HBr is a strong acid. It dissociates completely in water.
Most hydroxides are only slightly soluble. Unless the solution is quite dilute, I would write their states in water as "(s)", i.e., a suspension of the solid in water.
Salts containing Br⁻ are generally soluble.

Acids and bases react to give salts and water.

Thus, the molecular equation is

2HBr(aq) + Sr(OH)₂(s) ⟶ SrBr₂(aq) + 2H₂O(ℓ)

B. Ionic equation

You write all the soluble substances as ions.

2H⁺(aq)+ 2Br⁻(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2Br⁻(aq) + 2H₂O(ℓ)

C. Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

2H⁺(aq) + 2Br⁻(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2Br⁻(aq) + 2H₂O(ℓ)

The net ionic equation is

2H⁺(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2H₂O(ℓ)

3 0

2 years ago

I need some help understanding this.​

I need some help understanding this.