1.) Watch the animation, and observe the titration process using a standard 0.10
ID: 917529 • Letter: 1
Question
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)NaCl(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)×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
Explanation / Answer
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 : Before equivalence point [As before equivalence the concentration of H+ is greater than the concentration of OH- ]
B.) The pH of the solution is close to 12 : After equivalence point [As after equivalence the concentration of OH- is greater than the concentration of H+]
C.) The pH of the solution is 7 : At equivalence [As at equivalence the H+ = concentration of [OH-]
D.) The color of the solution is pink : AFter equivalence point as after equivalence point [OH-] is high and the phenophthalein show pink colour in basic medium.
E.) The color of the solution is blue: Before equivalence point
F.) The pH changes most rapidly: Just after equivalence point
2.) Consider the balanced equation for the titration of an acid, HCl, with a base, NaOH:
HCl(aq)+NaOH(aq)?NaCl(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)×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
For complete neturalization of HCl we need equal moles of NaOH
(Molarity X volume )acid = (Molarity x volume ) base
1 X 100 = 1 X Volume of base
Volume of base = 100mL [the volume to reach equivalence point]
A.) 50.0 mL of 1.00 M NaOH
So this is before equivalence point
B.) 150 mL of 1.00 M NaOH
This is after equivalence point
C.) 200 mL of 1.00 M NaOH
This if after equivalence point
D.) 10.0 mL of 1.00 M NaOH
This is before equivalence point
E.) 5.00 mL of 1.00 M NaOH
This is before equivalence point
F.) 100 mL of 1.00 M NaOH
This is point of equivalence point
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