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Please write a paragraph of the objective stated using the two images attached below Please please answer as fast as possible thank you

Please write a paragraph of the objective stated using the two images attached below Please please answer as fast as possible thank you

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please write a paragraph of the objective stated using the two images attached below Please please answer as fast as possible thank you
EXPERIMENT 8 pK AND EQUIVALENT WEIGHT BY POTENTIOMETRIC TITRATION OF AN ACID PURPOSE In this experiment you are to determine the pK and equivalent weight of a weak monoprotic acid by potentiometric titration. METHOD For an acid, the equivalent weight is defined as the mass in grams that provides one mole of protons in a neutralization reaction. Thus, for a polyprotic acid, the equivalent weight varies according to the neutralization reaction occurring and the amount of base allowed to react with the acid. For a monoprotic acid, HA, one mole of acid will provide one mole of protons, and therefore the equivalent weight equals the molar mass. Note that this is only true because the acid is monoprotic! When such an acid reacts with a strong base, the net ionic equation is: HA(aq) + OH(aq) A (aq) + H₂0 (1) In this experiment, you will weigh out a definite mass of a solid monoprotic acid. This amount of acid will be reacted with a solution of the strong base, sodium hydroxide, whose molarity is known. By determining the volume of NaOH needed to react completely with HA following Equation (1), you can calculate the molar mass of the weak acid. The volume of base needed for complete reaction will be determined by the method called titration. In this titration small, measured amounts of the NaOH solution are added to the solution of HA. The point in the titration at which the moles of added base just equal the moles of added acid is called the equivalence point. The volume of NaOH added at the equivalence point, Vap (measured to 0.01mL), can be used to calculate the equivalent weight (in this case, the molar mass) of the acid. ер, Several methods can be used to detect the equivalence point. In this experiment we will follow the pH of the solution as the base is added. A typical titration curve is shown in figure 1. Notice that at the equivalent point the pH "jumps." A mathematical analysis of the data collected in this experiment will allow the determination of this jump and thus of the equivalence point volume. This mathematical analysis will be done by computer. Fig. 1: pH vs vol. of NaOH for the titration of a weak monoprotic acid by a strong base. 1 V (mL NaOH)
Transcribed Image Text:EXPERIMENT 8 pK AND EQUIVALENT WEIGHT BY POTENTIOMETRIC TITRATION OF AN ACID PURPOSE In this experiment you are to determine the pK and equivalent weight of a weak monoprotic acid by potentiometric titration. METHOD For an acid, the equivalent weight is defined as the mass in grams that provides one mole of protons in a neutralization reaction. Thus, for a polyprotic acid, the equivalent weight varies according to the neutralization reaction occurring and the amount of base allowed to react with the acid. For a monoprotic acid, HA, one mole of acid will provide one mole of protons, and therefore the equivalent weight equals the molar mass. Note that this is only true because the acid is monoprotic! When such an acid reacts with a strong base, the net ionic equation is: HA(aq) + OH(aq) A (aq) + H₂0 (1) In this experiment, you will weigh out a definite mass of a solid monoprotic acid. This amount of acid will be reacted with a solution of the strong base, sodium hydroxide, whose molarity is known. By determining the volume of NaOH needed to react completely with HA following Equation (1), you can calculate the molar mass of the weak acid. The volume of base needed for complete reaction will be determined by the method called titration. In this titration small, measured amounts of the NaOH solution are added to the solution of HA. The point in the titration at which the moles of added base just equal the moles of added acid is called the equivalence point. The volume of NaOH added at the equivalence point, Vap (measured to 0.01mL), can be used to calculate the equivalent weight (in this case, the molar mass) of the acid. ер, Several methods can be used to detect the equivalence point. In this experiment we will follow the pH of the solution as the base is added. A typical titration curve is shown in figure 1. Notice that at the equivalent point the pH "jumps." A mathematical analysis of the data collected in this experiment will allow the determination of this jump and thus of the equivalence point volume. This mathematical analysis will be done by computer. Fig. 1: pH vs vol. of NaOH for the titration of a weak monoprotic acid by a strong base. 1 V (mL NaOH)
The pKa of the acid can also be deduced from the titration data. Ka, the acid dissociation constant, is the equilibrium constant for the dissociation of the acid in water. HA(aq) + H20 (1) H30* (aq) + A¯ (aq) (1) Ka= [H30+][A] [HA] The pKa of the acid is the negative of the base 10 logarithm of the Ka. pKa = -log10 Ka (2) (3) The pH of a solution of a weak acid is related to the pK of the acid by the equation: [A] pH = pKa + log[HA] (4) In a solution for which [A] = [HA], pH = pKa [Can you derive this from equation (4)?] In this experiment, OH¯ equal one-half the equivalents of HA initially present, [A] = [HA] and thus pH = pKa [This point in the titration is known as the buffer point; the position of the buffer point in the titration curve is shown on Figure 1.] The volume needed to reach the buffer point, Vbp,, will be one-half that needed to reach the equivalence point. Vbp = 1/2 Vep (5) In summary, the titration will be performed and data collected for pH versus volume of NaOH added. By a mathematical analysis of the titration curve, the equivalence point volume will be identified. From this volume, the equivalent weight can be calculated. Again by mathematical analysis, the pH at the buffer point will be determined. This is a measure of pKa of the acid.
Transcribed Image Text:The pKa of the acid can also be deduced from the titration data. Ka, the acid dissociation constant, is the equilibrium constant for the dissociation of the acid in water. HA(aq) + H20 (1) H30* (aq) + A¯ (aq) (1) Ka= [H30+][A] [HA] The pKa of the acid is the negative of the base 10 logarithm of the Ka. pKa = -log10 Ka (2) (3) The pH of a solution of a weak acid is related to the pK of the acid by the equation: [A] pH = pKa + log[HA] (4) In a solution for which [A] = [HA], pH = pKa [Can you derive this from equation (4)?] In this experiment, OH¯ equal one-half the equivalents of HA initially present, [A] = [HA] and thus pH = pKa [This point in the titration is known as the buffer point; the position of the buffer point in the titration curve is shown on Figure 1.] The volume needed to reach the buffer point, Vbp,, will be one-half that needed to reach the equivalence point. Vbp = 1/2 Vep (5) In summary, the titration will be performed and data collected for pH versus volume of NaOH added. By a mathematical analysis of the titration curve, the equivalence point volume will be identified. From this volume, the equivalent weight can be calculated. Again by mathematical analysis, the pH at the buffer point will be determined. This is a measure of pKa of the acid.
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