# Determine the concentration of the original NaOH solution

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.

• user3764899 - good answer. But what a strange question. Why didn't the chemist just titrate the $\ce{NaOH}$ with the nickel directly? If the concentration of $\ce{HCl}$ were unknown and that of $\ce{NaOH}$ were known, the procedure described would make more sense, but as it is, it makes me wonder whether this teacher ever did any chemistry. – Silvio Levy Jul 25 '14 at 22:22