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Chapter 17: Problem 42

How many milliliters of \(0.105 \mathrm{MHCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 \(\mathrm{mL}\) of \(0.0950 \mathrm{MNaOH}\) (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{MNH}_{3}\), (c) 125.0 mL of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Short Answer

Expert verified
To titrate each solution mentioned to the equivalence point, the following volumes of 0.105 M HCl are needed: (a) 40.71 mL for 45.0 mL of 0.0950 M NaOH. (b) 25.29 mL for 22.5 mL of 0.118 M NH3. (c) 40.18 mL for 125.0 mL of a solution that contains 1.35 g of NaOH per liter.

Step by step solution

01

Determine moles of NaOH

To find the moles of NaOH, we will multiply its molarity (0.0950 M) by its volume in liters (0.0450 L): Moles of NaOH = (0.0950 mol/L) * (0.0450 L) = 0.004275 mol
02

Find moles of HCl reactant

Since the molar ratio of NaOH to HCl is 1:1 in their reaction, we need an equal amount of moles of HCl to titrate the NaOH: Moles of HCl = 0.004275 mol
03

Calculate the volume of 0.105 M HCl

Now, we'll divide the moles of HCl by its molarity to find the required volume to titrate the NaOH: Volume of HCl = (0.004275 mol) / (0.105 mol/L) = 0.040714 L = 40.71 mL Answer for problem (a): 40.71 mL of 0.105 M HCl. ### Problem (b): 22.5 mL of 0.118 M NH3 ###
04

Determine moles of NH3

Moles of NH3 = (0.118 mol/L) * (0.0225 L) = 0.002655 mol
05

Find moles of HCl reactant

The molar ratio of NH3 to HCl is 1:1, therefore: Moles of HCl = 0.002655 mol
06

Calculate the volume of 0.105 M HCl

Volume of HCl = (0.002655 mol) / (0.105 mol/L) = 0.025286 L = 25.29 mL Answer for problem (b): 25.29 mL of 0.105 M HCl. ### Problem (c): 125 mL of a solution containing 1.35 g of NaOH per liter ###
07

Calculate moles of NaOH in the solution

First, find the molarity of the NaOH solution: 1.35 g/L * (1 mol/39.997 g) = 0.033749 M (we multiply by the inverse of NaOH's molar mass to convert grams to moles). Now, determine moles of NaOH in our given volume: Moles of NaOH = (0.033749 mol/L) * (0.125 L) = 0.0042186 mol
08

Find moles of HCl reactant

Moles of HCl = 0.0042186 mol (1:1 molar ratio of NaOH to HCl)
09

Calculate the volume of 0.105 M HCl

Volume of HCl = (0.0042186 mol) / (0.105 mol/L) = 0.0401771 L = 40.18 mL Answer for problem (c): 40.18 mL of 0.105 M HCl.

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Most popular questions from this chapter

A solution contains three anions with the following concentrations: $0.20 \mathrm{M} \mathrm{CrO}_{4}^{2-}, 0.10 \mathrm{M} \mathrm{CO}_{3}^{2-},$ and \(0.010 \mathrm{M} \mathrm{Cl}^{-}\). If a dilute \(\mathrm{AgNO}_{3}\) solution is slowly added to the solution, what is the first compound to precipitate: $\mathrm{Ag}_{2} \mathrm{CrO}_{4}\left(K_{s p}=1.2 \times 10^{-12}\right), \mathrm{Ag}_{2} \mathrm{CO}_{3}\left(K_{s p}=8.1 \times 10^{-12}\right)$ or \(\operatorname{AgCl}\left(K_{s p}=1.8 \times 10^{-10}\right) ?\)

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{pK}_{b}\) of its base component.

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