CHE-1102 Lecture Notes - Lecture 7: Titration Curve, Equivalence Point, Sodium Hydroxide

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9 Nov 2017

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Acid base equilibrium titrations with: titration a procedure for determining the concentration of a solution using another solution of. Base known concentration called a standard solution: equivalence point data point on the titration curve where the number of moles of acid. = number of moles of base: titration a 25. 0ml sample of 0. 300 m hcl is titrated with 0. 150 m naoh. Initial ph ph = -log (0. 300) = 0. 523 (volume of naoh added = 25. 0 ml) Initial moles of acid = (0. 0250l)(0. 300m) = 0. 00750 mol hcl. Moles of base added = (0. 0250l)(0. 150m) = 0. 00375 mol naoh. Enter data in this table as moles. ph = 7. 00. 25. 0 ml + 25. 0 ml = 50. 0 ml total. Calculate the ph of the solution. ph = -log (0. 075) = 1. 12. Naoh added to hcl : ha + oh- . The equivalence point has a ph > 7 (basic: titration 25. 0 ml of 0. 16 m acetic acid is titrated with.

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

Textbook is showing HCl + NaOH titrating problem. After adding 5 mL of NaOH, why is the equation OH + H3O _--> 2h2o? Where is the h3o coming from

Ammonia (NH₃) is a weak base with a pKb of 4.75. Its conjugate acid, ammonium (NH₄⁺) has a pKa of 9.25. Set up a titration using 20.00 mL of HCL to titrate 20.00 mL of aqueous ammonia solution.

Concentration of HCl soln: 0.1044 mol/L

Initial buret reading (mL): 0.00

Buret reading at equivalence pt (mL): 14.41

Calculate the following:

a.) Moles of HCl added at equivalence pt

b.) Moles of NH₃ in sample

c.) NH₃ concentration (mol/L)

d.) pH at equivalence point

e.) pH at half the volume required to reach equivalence

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CHE-1102 Lecture Notes - Lecture 7: Titration Curve, Equivalence Point, Sodium Hydroxide

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