Question

Write a balanced molecular equation for a solution of nickel (II) chloride being added to a solution of sodium hydroxide, forming a precipitate. 1NiCl2(aq) + 2NaOH(aq) à 1Ni(OH)2(s) + 2NaCl(aq) (ii) What is the balanced net ionic equation piece of for the reaction in part (i) where a solution of nickel(II) chloride is added to a solution of sodium hydroxide, forming a precipitate. 1Ni+2(aq) + 2OH-1(aq) à 1Ni(OH)2(s) (iii)If equal volumes of 1.0 M nickel (II) chloride and 1.0 M sodium hydroxide are used, what ion is present in the solution in the highest concentration after the precipitate forms?

Answer 1

The balanced molecular equation is 1NiCl2(aq) + 2NaOH(aq) → 1Ni(OH)2(s) + 2NaCl(aq). The balanced net ionic equation is 1Ni+2(aq) + 2OH-1(aq) → 1Ni(OH)2(s). The ion present in the highest concentration after the precipitate forms is Nickel(II) ions (Ni+2).

Step-by-step explanation:

The balanced molecular equation for the reaction where a solution of nickel(II) chloride is added to a solution of sodium hydroxide, forming a precipitate is:

1NiCl2(aq) + 2NaOH(aq) → 1Ni(OH)2(s) + 2NaCl(aq)

The balanced net ionic equation for the same reaction is:

1Ni+2(aq) + 2OH-(aq) → 1Ni(OH)2(s)

After the precipitate forms, the ion present in the highest concentration is Nickel(II) ions (Ni+2).

Answer 2

Step-by-step explanation:

i - NiCl2(aq) + 2NaOH(aq) --> Ni(OH)2(s) + 2NaCl(aq)

ii - Ni2+(aq) + 2OH-(aq) --> Ni(OH)2(s)

iii -

For each mole of NiCl2 we need 2 moles of NaOH, for 1 mole of each we have the following case at the start:

NiCl2(aq) + 2NaOH(aq) --> Ni(OH)2(s) + 2NaCl(aq)

1 M 1M 0 0

At the end of the reaction we have in the solution:

NiCl2(aq) + 2NaOH(aq) --> Ni(OH)2(s) + 2NaCl(aq)

0,5M 0 0 1M

Ni(OH)2 is 0 because it's a precipitate, therefore it's not on the solution, but it should have 1M.

So the ion present in the biggest concentration should be Cl-

Because we have 1 M form the NaCl and 1 M for the NiCl2.