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.