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:


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)�~volume (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

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asked by LuigiR
  1. If I knew how to help I would but you don't have any questions here without referring to some web site that isn't listed.
    At the equivalence point the pH will be 7 for both titrations.
    Before the equivalence point for each, determine mols of HCl and mols NaOH from mols = M x L = ?
    If HCl is in excess mols, then (H^+) = M HCl = excess mols HCl/L and pH from there.
    If NaOH is in excess mols, then (OH^-) = excess mols NaOH and M NaOH at that point is M NaOH = mols excess NaOH/L solution and pOH and pH from there.

    After the equivalence point, the process is the same as before the eq point. Find the excess HCl or excess NaOH, M = mols acid or base/L solution and go to pH.

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  2. e,d,a,f,b,c

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    posted by latavia

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