CHEM 1133 Lecture 10: Chemistry 2-20

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20 Feb 2017

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Part 1: consider the reaction between hcl (aq) and naoh (aq: hcl (aq) + naoh (aq) h2o (l) + nacl (aq) (cid:1865)(cid:1867)(cid:1864)(cid:1857)=(cid:882). (cid:883) (cid:1839) (cid:882). (cid:882)(cid:884)(cid:887) (cid:1838)=(cid:882). (cid:882)(cid:882)(cid:884)(cid:887) (cid:1865)(cid:1867)(cid:1864) (cid:1864) The reaction between an acid and a base is called a neutralization reaction: we can apply the principles of acid-base neutralization and stoichiometry to a common lab procedure called a titration. In a titration, a substance in a solution of a known concentration is reacted with another substance in a solution of unknown concentration. The equivalence point is when acid = base. (cid:4666)(cid:884)(cid:887) (cid:1865)(cid:1838)(cid:4667)(cid:4666)(cid:882). (cid:882)(cid:887) (cid:1839)(cid:4667)=(cid:882). (cid:882)(cid:882)(cid:883)(cid:884)(cid:887) (cid:1840)(cid:1841) limiting reactant. (cid:4666)(cid:884)(cid:887) (cid:1865)(cid:1838)(cid:4667)(cid:4666)(cid:882). (cid:883) (cid:1839)(cid:4667)= (cid:882). (cid:882)(cid:882)(cid:884)(cid:887) (cid:1864) excess reactant. (cid:882). (cid:882)(cid:887) (cid:1839) (cid:882). (cid:882)(cid:887) (cid:1838) = (cid:882). (cid:882)(cid:882)(cid:884)(cid:887) (cid:1865)(cid:1867)(cid:1864) (cid:1840)(cid:1841) (cid:882). (cid:882)(cid:882)(cid:884)(cid:887) (cid:1865)(cid:1867)(cid:1864) (cid:1864) (cid:882). (cid:882)(cid:882)(cid:884)(cid:887) (cid:1865)(cid:1867)(cid:1864) = (cid:882) (cid:1865)(cid:1867)(cid:1864) (cid:1864) (cid:1864)(cid:1857)(cid:1858) equivalence point. All hcl used up after the addition of 50 ml of naoh. H2o oh- + h: kw=10-14.

Related Questions

1.) Watch the animation, and observe the titration process using a standard 0.100 M sodium hydroxide solution to titrate 50.0 mL of a 0.100 M hydrochloric acid solution. Identify which of the following statements regarding acid-base titrations are correct

Drag items A-F to: Before equivalence point, At equivalence point, After equivalence point

A.) The pH of the solution is close to 2

B.) The pH of the solution is close to 12

C.) The pH of the solution is 7

D.) The color of the solution is pink

E.) The color of the solution is blue

F.) The pH changes most rapidly

2.) Consider the balanced equation for the titration of an acid, HCl, with a base, NaOH:

HCl(aq)+NaOH(aq)âNaCl(aq)+H2O(aq)

According to the balanced equation, to neutralize one mole of the acid (HCl) you will have to add one mole of the base (NaOH). Based on the number of moles of HCl and NaOH at a particular stage of titration, the progress of the reaction with respect to the equivalence point can be decided, as shown in the following table:

You can use the following equation to determine the number of moles in a given solution or volume of titrant: number of moles=molarity (mol/L)Ãvolume (in liters)

A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions based on whether they are before the equivalence point, at the equivalence point, or after the equivalence point.

Drag items A-F to Before equivalence point, At equivalence point, After equivalence point

A.) 50.0 mL of 1.00 M NaOH

B.) 150 mL of 1.00 M NaOH

C.) 200 mL of 1.00 M NaOH

D.) 10.0 mL of 1.00 M NaOH

E.) 5.00 mL of 1.00 M NaOH

F.) 100 mL of 1.00 M NaOH

A titration is a procedure for determining the concentration of a solution by allowing it to react with another solution of known concentration (called a standard solution). Acid-base reactions and oxidation-reduction reactions are used in titrations. For example, to find the concentration of an HCl solution (an acid), a standard solution of NaOH (a base) is added to a measured volume of HCl from a calibrated tube called a buret. An indicator is also present and it will change color when all the acid has reacted. Using the concentration of the standard solution and the volume dispensed, we can calculate molarity of the HCl solution.

A) A volume of 90.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 14.7 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(aq)+H2SO4(aq)âK2SO4(aq)+2H2O(l)

B) Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydrogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction:

2KMnO4(aq)+H2O2(aq)+3H2SO4(aq)â3O2(g)+2MnSO4(aq)+K2SO4(aq)+4H2O(l)

A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.68 M KMnO4. What mass of H2O2 was dissolved if the titration required 18.8 mL of the KMnO4 solution?

CHEM 1133 Lecture 10: Chemistry 2-20

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