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Chapter 4: Problem 127

What volume of 0.100\(M \mathrm{NaOH}\) is required to precipitate all of the nickel(Il) ions from 150.0 \(\mathrm{mL}\) of a \(0.249-M\) solution of \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short Answer

Expert verified
The volume of 0.100 M NaOH required to precipitate all of the nickel(II) ions from 150.0 mL of 0.249 M Ni(NO3)2 solution is 747 mL.

Step by step solution

01

Equation

The precipitate formation occurs when NaOH reacts with Ni(NO3)2 to form solid Ni(OH)2 and NaNO3. The balanced chemical equation is: \(Ni(NO_3)_2 + 2NaOH \rightarrow Ni(OH)_2 + 2NaNO_3\) Step 2: Convert the volume and concentration of Ni(NO3)2 to moles
02

Moles of Ni(NO3)2

Using the given volume and concentration of Ni(NO3)2, we can find the moles of Ni(NO3)2: moles = concentration × volume moles of \(Ni(NO_3)_2 = 0.249M \times 150.0mL \times \frac{1L}{1000mL}\) moles of \(Ni(NO_3)_2 = 0.03735 mol\) Step 3: Find the moles of NaOH needed
03

Moles of NaOH

According to the balanced chemical equation in step 1, 2 moles of NaOH is needed for every mole of Ni(NO3)2. So, we can calculate the moles of NaOH: moles of NaOH = moles of \(Ni(NO_3)_2 \times \frac{2 \, moles \, of\, NaOH}{1\, mole\, of\, Ni(NO_3)_2}\) moles of NaOH = 0.03735 mol × 2 moles of NaOH = 0.0747 mol Step 4: Calculate the volume of NaOH needed
04

Volume of NaOH

Now that we know the moles of NaOH needed to precipitate all the nickel(II) ions, we can find the volume of 0.100 M NaOH solution required: volume = moles / concentration volume of NaOH = 0.0747 mol / 0.100 M volume of NaOH = 0.747 L Step 5: Convert the volume of NaOH to the desired unit
05

Final Volume

We'll convert the volume of NaOH from liters to milliliters: volume of NaOH = 0.747 L × \(1000 \frac{mL}{L}\) volume of NaOH = 747 mL The volume of 0.100 M NaOH required to precipitate all of the nickel(II) ions from 150.0 mL of 0.249 M Ni(NO3)2 solution is 747 mL.

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Most popular questions from this chapter

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