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Chapter 4: Problem 113

What volume of \(0.100 M \mathrm{NaOH}\) is required to precipitate all of the nickel(II) 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
747 mL of 0.100 M NaOH is required to precipitate all of the nickel(II) ions from 150.0 mL of a 0.249-M solution of \(Ni(NO_3)_2\).

Step by step solution

01

1. Write the balanced chemical equation

First, let's write the balanced chemical equation for the reaction between NaOH and Ni(NO3)2: \(Ni(NO_{3})_{2}(aq) + 2 NaOH(aq) \rightarrow Ni(OH)_{2}(s) + 2 NaNO_{3}(aq)\) Here, the stoichiometric ratio between Ni(NO3)2 and NaOH is 1:2, meaning that 1 mole of Ni(NO3)2 reacts with 2 moles of NaOH.
02

2. Find the number of moles of Ni(NO3)2

Now, we have to calculate the number of moles of Ni(NO3)2 in 150.0 mL of a 0.249-M solution. We can do this using the formula: moles = molarity × volume (in Liters). moles of Ni(NO3)2 = 0.249 mol/L × 0.150 L = 0.03735 mol Ni(NO3)2
03

3. Calculate the required number of moles of NaOH

Using the stoichiometric ratio from the balanced chemical equation, we can find the required number of moles of NaOH: moles of NaOH = 2 moles NaOH/1 mole Ni(NO3)2 × 0.03735 mol Ni(NO3)2 = 0.0747 mol NaOH
04

4. Calculate the required volume of NaOH

Finally, we can calculate the required volume of 0.100 M NaOH using the formula: volume (L) = moles / molarity. Volume of NaOH = 0.0747 mol NaOH / 0.100 mol/L = 0.747 L
05

5. Convert the volume to mL

To get the volume in milliliters, multiply the volume in liters by 1000: Volume of NaOH = 0.747 L × 1000 mL/L = 747 mL So, 747 mL of 0.100 M NaOH is required to precipitate all of the nickel(II) ions from 150.0 mL of a 0.249-M solution of Ni(NO3)2.

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Most popular questions from this chapter

What mass of solid aluminum hydroxide can be produced when \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is added to \(200.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{KOH} ?\)

A \(30.0-\mathrm{mL}\) sample of an unknown strong base is neutralized after the addition of \(12.0 \mathrm{~mL}\) of a \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is \(0.0300 M\), give some possible identities for the unknown base.

A \(450.0-\mathrm{mL}\) sample of a \(0.257-M\) solution of silver nitrate is mixed with \(400.0 \mathrm{~mL}\) of \(0.200 M\) calcium chloride. What is the concentration of \(\mathrm{Cl}^{-}\) in solution after the reaction is complete?

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