Nicl3 is a strong electrolyte. determine the concentration of each of the individual ions in a 0.450 m nicl3 solution

Nicl3 is a strong electrolyte. determine the concentration of each of the individual ions in a 0.450 m nicl3 solution

Answers

NiCl3 is a strong electrolyte. This means that when on solution it dissociates completely into ions. The dissociation reaction would be written as:

NiCl3 = Ni3+ + Cl-

The ratio of the ions and the compound are all 1 is to 1. Therefore, the concentration of the individual ions would be the same as the concentration of the compound at a certain volume.

The concentration of each of the individual ions in a 0750 M Ba(OH)2 solution is

[Ba2+] = 0.750 M

[OH-]= 1.50 M

calculation

write the equation for dissociation

that is  Ba(OH)2  (s)→ Ba2+(aq)  + 2OH-(aq)

by use of mole ratio of Ba(OH)2 : Ba2+ which is 1: 1 the concentration of Ba2+ is therefore= 0.750M

by use of mole ratio of Ba(OH)2 : OH- which is 1:2 the concentration of  OH- =0.750 M x2/1=1.50 M

Ba^+^2 = 0.3 M

OH^- = 0.6 M

Explanation:

We can start wit the ioniation reaction:

Ba(OH)_2~-~Ba^+^2~+~2OH^-

We have the ions Ba^+^2 and OH^-. Now, we can calculate the concentration of each ion:

Concentration of Ba^+^2

In balanced reaction we have 1 mol of Ba(OH)_2 and 1 mol of Ba^+^2, so, we have a 1:1 mol ratio, with this in mind:

0.300~M~Ba(OH)_2frac{1~Ba^+^2}{1~Ba(OH)_2}=0.300~M~Ba^+^2

Concentration of OH^-

In balanced reaction we have 1 mol of Ba(OH)_2 and 2 mol of OH^-, so, we have a 1:2 mol ratio, with this in mind:

0.300~M~Ba(OH)_2frac{2~OH^-}{1~Ba(OH)_2}=0.600~M~OH^-

I hope it helps!

Answer : The concentration of each of the individual ions are, 0.6 M Ni^{3+} and 1.8 M Cl^-.

Explanation : Given,

Concentration of NiCl_3 solution = 0.6 M

The balanced chemical reaction will be :

NiCl_3rightarrow Ni^{3+}+3Cl^-

From the balanced chemical reaction we conclude that, 1 mole of NiCl_3 dissociates to give 1 mole of Ni^{3+} and 3 moles of Cl^-.

So,

The concentration of Ni^{3+} = The concentration of NiCl_3 = 0.6 M

The concentration of Cl^- =  3 × The concentration of NiCl_3 = 3times 0.6M=1.8M

Therefore, the concentration of each of the individual ions are, 0.6 M Ni^{3+} and 1.8 M Cl^-.

Answer;

[Ba2+] = 0.550 M  

[OH-] = 1.10 M

Explanation;

Each mol Ba(OH)2 releases 2 mol OH-  

Ba(OH)2 > Ba2+ + 2OH-  

Mole ratio; Ba(OH)2 :Ba2+ = 1 : 1

[Ba2+] = 0.550 M  

Mole ratio; Ba(OH)2: 2OH-  = 1 : 2

= 2 × 0.550 = 1.10

[OH-] = 1.10 M  

This assumes complete ionization (dissociation) of Ba(OH)2.

[Ba^2+]= 0.400M

[OH-]= 0.800M

Explanation:

Step 1: Data given

Molarity ofBa(OH)2 = 0.400 M

Ba(OH)2 is a strong electrolyte, so 100 % dissociation

Step 2: The equation

Ba(OH)2 → Ba^2+ +  2OH-

In 1 mol Ba(OH)2 we have 1 mol Ba^2+ and 2 moles OH-

Step 3: Calculate the concentration

[Ba^2+]= 1* 0.400 M = 0.400M

[OH-]= 2*0.400 M = 0.800M

A strong electrolyte is a solute that completely, or almost completely, ionizes or dissociates in a solution. Therefore, NiCl3 dissociates into ions completely. The dissociatio equation would be:

NiCl3 ---> Ni3+ + 3Cl-1

Threfore, the concentration of the ions would be 0.4 M Ni3+ and 1.2 M Cl-1.

Conc. of K+ ions = 0.90 MCoc. of SO4∧-2 = 0.45 M

Explanation:

Data Given:

Conc. of H2SO4 = 0.450

As sulphoric acid is a strong electrolyte, it completely dissociate into ions.

H2SO4 ⇆   2K+   +   SO4∧-2

.450 M K2SO4 means that there is .450 mols of K2SO4 in every liter of solution.

                      K2SO4  :  K+                                          K2SO4  : SO4∧-2

                          1      =     2                                                   1     =    1

                     0.450   =   2× 0.450 = 0.90                     0.450 =  0.450×1 = 0.450

Result:

Conc. of potassium ion will be 0.90M    

Coc. of sulphate ions will be 0.45 M

For the answer to the question above, I'll show the solutions for my answer.
(0.800 M NiCl3) x (1 mol Ni^3+ / 1 mol NiCl3) = 0.800 M Ni^3+ 
(0.800 M NiCl3) x (3 mol Cl^- / 1 mol NiCl3) = 2.40 M Cl^-
I hope my answer helped you. Have a nice day!
Potassium sulfate or K₂SO₄ is a strong electrolyte which means that it will completely dissociate into ions when dissolved in water. The dissociation reaction is as follows:

K₂SO₄ → 2K⁺ + SO₄²⁻ 

This means that the dissociation of 1 mol of K₂SO₄ solution will yield 2 moles of K⁺ and 1 mole of SO₄²⁻. Molarity is a measure of concentration defined as moles divided by the volume of solution. In this case, given that the volume is constant, from stoichiometry, the concentration of K⁺ ions would be twice of 0.600 M which is 1.200 M. On the other hand, the concentration of SO₄²⁻ is 0.600 M by stoichiometry.    

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