Determine the volume (ml) of 0.1 N NaOH solution that will be required for the complete precipitation of nickel

Determine the volume (ml) of 0.1 N NaOH solution that will be required for the complete precipitation of nickel ions from 100 ml of a 20% solution of Ni (No3) 2 with a density of 1.191 g / cm3

Decision:
M [Ni (No3) 2] = 182.7 g / mol.
The reaction equation has the form:
Ni (No3) 2 + 2NaOH = 2NaNO3 + Ni (OH) 2 ↓
From the reaction equation it follows that 1 mol of Ni (No3) 2 reacts with 2 mol of NaOH.
From here
n [(Ni (No3) 2] = 2n (NaOH).
We calculate the mass of Ni (No3) 2 spent on the reaction, we obtain:
m [Ni (No3) 2] = [w%. V [Ni (No3) 2]. p (p-pa)] 100% = (20. 100. 1,191) / 100 = 23.82 g.
We find the amount of Ni (No3) 2, we get:
n [Ni (No3) 2] = m [Ni (No3) 2] / M [Ni (No3) 2] = 23.82 / 182.7 = 0.13 mol.
From here
n (NaOH) = 2n [Ni (No3) 2] = 2. 0.13 = 0.26 mol.
We calculate the volume of NaOH solution, which is required for the complete deposition of nickel ions, we obtain:
V (NaOH) = [n (NaOH). 1000] / CH (NaOH) = (0.26. 1000) / 0.1 = 2600 ml.
Answer: V (NaOH) = 2600 ml

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