Determine the concentration of the original NaOH solution

Question

This was a question on the mt exam two days ago :

A chemist added 40.0 mL of an $NaOH$ solution to 90.0 mL of 0.400 M $HCl$.

The solution was then treated with an excess of $\text {nickel(II) nitrate}$, resulting in the formation of 1.06 of $Ni(OH)_2$ precipitate. Determine the concentration of the original $NaOH$ solution.

I didn't know what to do with this question

I knew that I should get the number of mol of NaOH to get the solution's concentration

Then I wrote first the equation that describes the first part of the reaction :

$NaOH + HCl \rightarrow NaCl + H2O $

After that I wrote the second part of the reaction :

$NaCl + H_2O + Ni(NO_3)_2 \rightarrow Ni(OH)_2 + ? + ?$

I don't know what will be the the other products ?

I stopped here and I guessed an answer.

Answer 1

I am going to need a unit for the 1.06. The question needs you to figure out the amount in moles of Nickel Hydroxide. Then multiply this by two as two moles of NaOH are required to form one mole of the Nickel precipitate. Then this is the number of moles that was surplus in the reaction with HCl (NaOH is added in excess - the key to the question). Find the number of moles of HCl using the volume and concentration. Add this number to the number of moles that were used to form the Nickel precipitate and then find the concentration using the moles and the volume.