Can someone please help me

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]

<u>Answer:</u> The concentration of $NH_3$ required will be 0.285 M.

<u>Explanation:</u>

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

3 years ago

The position of equilibrium would not be appreciably affected by changes in the volume of the container for

8090 [49]

Answer:

The position of equilibrium would not be appreciably affected by changes in the volume of the container for NiO(s) + CO(g) ⇌ Ni(s) + CO2(g).

Correct Answer : Option A

Explanation:

The equilibrium position tends to change with increase or decrease in pressure or volume, or both of them. This happens because considering change in volume, when the volume of the container increases, the reactant molecule increases i.e. mole of gases and thus the position of equilibrium shifts towards the right side. Similarly in case of decreasing volume, reactant molecule decreases and the equilibrium position shifts left side.

And in case, when the mole of gases on both the sides of the equation i.e. reactant side and product side are equal, it will not have any effect on the equilibrium position on increasing or decreasing volume, or pressure.

4 0

3 years ago

What makes bullet holes wider on the exit side of a piece of glass?

denpristay [2]

The rounded shape of the bullet because as it goes through the glass it stops spinning and this make the hole wider

6 0

3 years ago

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Assume that you are given a solution of an unknown acid or base. How can you tell whether the unknown substance is acidic or bas

Jlenok [28]

Answer:

The litmus test.