. A student 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 nickel(II) nitrate, resulting in the formation
of 1.06 g of Ni(OH)2 precipitate. Determine the concentration of the original NaOH
solution.
To determine the concentration of the original NaOH solution, we can use stoichiometry to relate the amount of Ni(OH)2 precipitate formed to the amount of NaOH reacted.
First, let's write the balanced equation for the reaction between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
We can see that the ratio of NaOH to Ni(OH)2 is 1:1 based on the balanced equation.
Next, we need to calculate the moles of Ni(OH)2 precipitate formed. We know the mass of the precipitate is 1.06 g, so we need to convert that to moles. The molar mass of Ni(OH)2 is 92.71 g/mol.
moles of Ni(OH)2 = mass of Ni(OH)2 / molar mass of Ni(OH)2
moles of Ni(OH)2 = 1.06 g / 92.71 g/mol
moles of Ni(OH)2 = 0.0114 mol
Since the ratio of NaOH to Ni(OH)2 is 1:1, the moles of NaOH reacted is also 0.0114 mol.
Now, let's calculate the volume in liters of the NaOH solution used. The total volume of the solution is the sum of the volumes of NaOH and HCl.
Total volume = volume of NaOH + volume of HCl
Total volume = 40.0 mL + 90.0 mL
Total volume = 130.0 mL = 0.130 L
With the moles of NaOH (0.0114 mol) and the volume of NaOH solution (0.130 L), we can calculate the concentration of the NaOH solution using the formula:
Concentration = moles / volume
Concentration = 0.0114 mol / 0.130 L
Concentration of the original NaOH solution = 0.0877 M