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Chapter 17: Problem 41

How many milliliters of \(0.0850 \mathrm{M} \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3}\), (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH},\) (c) \(50.0 \mathrm{~mL}\) of a solution that contains \(1.85 \mathrm{~g}\) of HCl per liter?

Short Answer

Expert verified
The required volumes of 0.0850 M NaOH to titrate each solution to the equivalence point are: (a) 42.4 mL for 40.0 mL of 0.0900 M HNO3, (b) 35.0 mL for 35.0 mL of 0.0850 M CH3COOH, (c) 29.8 mL for 50.0 mL of a solution that contains 1.85 g of HCl per liter.

Step by step solution

01

(a) Moles of HNO3

To calculate the moles of HNO3, we will use the given volume (40 mL) and concentration (0.0900 M): Moles of HNO3 = Volume × Concentration Moles of HNO3 = \(40.0 \times 10^{-3} L\) × 0.0900 M = 0.0036 mol
02

(b) Moles of CH3COOH

To calculate the moles of CH3COOH, we will use the given volume (35 mL) and concentration (0.0850 M): Moles of CH3COOH = Volume × Concentration Moles of CH3COOH = \(35.0 \times 10^{-3} L\) × 0.0850 M = 0.002975 mol
03

(c) Moles of HCl

First, calculate the mass of HCl in the 50.0 mL solution: Mass of HCl = \(1.85 \frac{g}{L}\) × \(50.0 \times 10^{-3} L\) = 0.0925 g Next, convert the mass of HCl to moles by dividing by its molar mass (36.461 g/mol): Moles of HCl = \(\frac{0.0925}{36.461}\) = 0.002536 mol ##Step 2: Determine the volume of NaOH required## Since the equivalence point is when the moles of acid equal the moles of the base, we can use the given concentration of NaOH to find the required volume for each scenario.
04

(a) Volume of NaOH for HNO3 titration

Using the moles of HNO3 calculated previously, we can determine the required volume of NaOH: Volume = \(\frac{Moles}{Concentration}\) Volume = \(\frac{0.0036}{0.0850}\) L = 0.0424 L = 42.4 mL
05

(b) Volume of NaOH for CH3COOH titration

Using the moles of CH3COOH calculated previously, we can determine the required volume of NaOH: Volume = \(\frac{Moles}{Concentration}\) Volume = \(\frac{0.002975}{0.0850}\) L = 0.0350 L = 35.0 mL
06

(c) Volume of NaOH for HCl titration

Using the moles of HCl calculated previously, we can determine the required volume of NaOH: Volume = \(\frac{Moles}{Concentration}\) Volume = \(\frac{0.002536}{0.0850}\) L = 0.0298 L = 29.8 mL So the required volumes of 0.0850 M NaOH to titrate each solution to the equivalence point are: (a) 42.4 mL for 40.0 mL of 0.0900 M HNO3, (b) 35.0 mL for 35.0 mL of 0.0850 M CH3COOH, (c) 29.8 mL for 50.0 mL of a solution that contains 1.85 g of HCl per liter.

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Most popular questions from this chapter

A buffer is prepared by adding \(20.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the \(\mathrm{pH}\) of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

(a) Explain the difference between solubility and solubilityproduct constant. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH}\) 7: (a) \(\mathrm{NaHCO}_{3}\) titrated with \(\mathrm{NaOH},\) (b) \(\mathrm{NH}_{3}\) titrated with \(\mathrm{HCl}\) (c) KOH titrated with HBr.

(a) Consider the equilibrium \(\mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) .\) Using Le Châtelier's principle, explain the effect of the presence of a salt of \(\mathrm{HB}^{+}\) on the ionization of B. (b) Give an example of a salt that can decrease the ionization of \(\mathrm{NH}_{3}\) in solution.

A sample of \(0.1687 \mathrm{~g}\) of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.1150 \mathrm{M} \mathrm{NaOH}\). The acid required \(15.5 \mathrm{~mL}\) of base to reach the equivalence point. (a) What is the molecular weight of the acid? (b) After \(7.25 \mathrm{~mL}\) of base had been added in the titration, the \(\mathrm{pH}\) was found to be 2.85 . What is the \(K_{a}\) for the unknown acid?
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