what volume of .1 M solution of NaOH is required to precipitate all of the nickel 2+ ions from 250 mL of a .273 M solution of nickel (II) nitrate?
To determine the volume of a 0.1 M NaOH solution required to precipitate all of the nickel 2+ ions from a 250 mL 0.273 M nickel (II) nitrate solution, we need to set up and solve an equation using the concept of stoichiometry.
The balanced chemical equation for the reaction between NaOH and nickel (II) nitrate is:
2 NaOH + Ni(NO3)2 → Ni(OH)2 + 2 NaNO3
From the balanced equation, we can see that the ratio of NaOH to Ni(NO3)2 is 2:1. This means that 2 moles of NaOH will react with 1 mole of Ni(NO3)2, resulting in the formation of 1 mole of Ni(OH)2.
First, let's calculate the number of moles of nickel (II) nitrate in the 250 mL solution. We can use the relation:
moles = concentration (M) × volume (L)
moles of Ni(NO3)2 = 0.273 M × 0.250 L
moles of Ni(NO3)2 = 0.06825 mol
According to the stoichiometry equation, 1 mole of Ni(NO3)2 reacts with 2 moles of NaOH. Therefore, we would need twice the number of moles of NaOH to react completely with 0.06825 moles of Ni(NO3)2.
moles of NaOH needed = 2 × 0.06825 mol
moles of NaOH needed = 0.1365 mol
Next, to find the volume of the 0.1 M NaOH solution required, we can use the relation:
volume (L) = moles / concentration (M)
volume of NaOH needed = 0.1365 mol / 0.1 M
volume of NaOH needed = 1.365 L
Finally, we need to convert the volume from liters to milliliters:
volume of NaOH needed = 1.365 L × 1000 mL/L
volume of NaOH needed = 1365 mL
Therefore, approximately 1365 mL, or 1.365 L, of a 0.1 M NaOH solution is required to precipitate all of the nickel 2+ ions from 250 mL of a 0.273 M solution of nickel (II) nitrate.