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Chapter 6: Problem 115

What volume of 0.100 \(M\) 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
747.0 mL of 0.100 M NaOH solution is required to precipitate all the nickel(II) ions from 150.0 mL of a 0.249 M Ni(NO₃)₂ solution.

Step by step solution

01

Write the balanced chemical equation for the reaction.

We want to know the reaction between NaOH and Ni(NO₃)₂: Ni(NO₃)₂ (aq) + 2 NaOH (aq) → Ni(OH)₂ (s) + 2 NaNO₃ (aq)
02

Find the number of moles of Ni(NO₃)₂.

Given that we have 150.0 mL of 0.249 M Ni(NO₃)₂, we can find the moles of \(Ni(NO_3)_2\). First, we need to convert the volume to liters, then multiply by the provided molarity: Moles of \(Ni(NO_3)_2\) = volume (in L) × molarity Moles of \(Ni(NO_3)_2\) = 0.150 L × 0.249 mol/L = 0.03735 mol
03

Calculate the moles of NaOH needed

Using the stoichiometry from the balanced chemical equation (1 Ni(NO₃)₂ requires 2 NaOH), we can find the number of moles of NaOH needed: Moles of NaOH = moles of \(Ni(NO_3)_2\) × 2 moles of NaOH / 1 mole of \(Ni(NO_3)_2\) Moles of NaOH = 0.03735 mol × 2 = 0.07470 mol
04

Determine the volume of NaOH needed

Now that we have the moles of NaOH needed, we can use the given molarity of NaOH to find the required volume (in L): Volume of NaOH = moles of NaOH ÷ molarity of NaOH Volume of NaOH = 0.07470 mol ÷ 0.1 M = 0.747 L Since we want the answer in mL, we'll multiply by 1000: Volume of NaOH = 0.747 L × 1000 = 747.0 mL Hence, 747.0 mL of the 0.100 M NaOH solution would be required to precipitate all the nickel(II) ions from 150.0 mL of the 0.249 M Ni(NO₃)₂ solution.

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Most popular questions from this chapter

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from \(75.0 \mathrm{mL}\) of a \(0.100-M\) solution of \(\mathrm{AgNO}_{3} ?\)

A 100.0-mL aliquot of 0.200 \(M\) aqueous potassium hydroxide is mixed with \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) aqueous magnesium nitrate. a. Write a balanced chemical equation for any reaction that occurs. b. What precipitate forms? c. What mass of precipitate is produced? d. Calculate the concentration of each ion remaining in solution after precipitation is complete.

If \(10 . \mathrm{g}\) of \(\mathrm{AgNO}_{3}\) is available, what volume of \(0.25 \mathrm{M} \mathrm{AgNO}_{3}\) solution can be prepared?

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing \(10.0,25.0,50.0,75.0,\) and \(100 .\) ppm of copper from a commercially produced 1000.0 -ppm solution? Assume each solution has a final volume of \(100.0 \mathrm{mL}\). (See Exercise 123 for definitions.)

What acid and what base would react in aqueous solution so that the following salts appear as products in the formula equation? Write the balanced formula equation for each reaction. a. potassium perchlorate b. cesium nitrate c. calcium iodide
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