# 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 Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.