Nicl3 is a strong electrolyte. determine the concentration of each of the individual ions in a 0.450 m nicl3 solution
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
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
= 0.3 M
= 0.6 M
We can start wit the ioniation reaction:
We have the ions and . Now, we can calculate the concentration of each ion:
In balanced reaction we have 1 mol of and 1 mol of , so, we have a 1:1 mol ratio, with this in mind:
In balanced reaction we have 1 mol of and 2 mol of , so, we have a 1:2 mol ratio, with this in mind:
I hope it helps!
Answer : The concentration of each of the individual ions are, 0.6 M and 1.8 M .
Explanation : Given,
Concentration of solution = 0.6 M
The balanced chemical reaction will be :
From the balanced chemical reaction we conclude that, 1 mole of dissociates to give 1 mole of and 3 moles of .
The concentration of = The concentration of = 0.6 M
The concentration of = 3 × The concentration of =
Therefore, the concentration of each of the individual ions are, 0.6 M and 1.8 M .
[Ba2+] = 0.550 M
[OH-] = 1.10 M
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.
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
NiCl3 ---> Ni3+ + 3Cl-1
Threfore, the concentration of the ions would be 0.4 M Ni3+ and 1.2 M Cl-1.
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
Conc. of potassium ion will be 0.90M
Coc. of sulphate ions will be 0.45 M
(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!
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.