Using the two images attached please write one paragraph explaining the analysis Also write another paragraph of the complete method in your own wordsPlease please please answer as fast as possible EXPERIMENT 8 pK AND EQUIVALENT WEIGHT BY POTENTIOMETRICTITRATION OF AN ACIDPURPOSEIn this experiment you are to determine the pK and equivalent weight of a weak monoprotic acidby potentiometric titration.METHODFor an acid, the equivalent weight is defined as the mass in grams that provides one mole ofprotons in a neutralization reaction. Thus, for a polyprotic acid, the equivalent weight variesaccording to the neutralization reaction occurring and the amount of base allowed to react withthe acid. For a monoprotic acid, HA, one mole of acid will provide one mole of protons, andtherefore the equivalent weight equals the molar mass. Note that this is only true because the acidis 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 ofacid will be reacted with a solution of the strong base, sodium hydroxide, whose molarity isknown. By determining the volume of NaOH needed to react completely with HA followingEquation (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 calledtitration. In this titration small, measured amounts of the NaOH solution are added to the solutionof HA. The point in the titration at which the moles of added base just equal the moles of addedacid 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 molarmass) of the acid.ер,Several methods can be used to detect the equivalence point. In this experiment we will followthe pH of the solution as the base is added. A typical titration curve is shown in figure 1. Noticethat at the equivalent point the pH "jumps." A mathematical analysis of the data collected in thisexperiment 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. ofNaOH for the titrationof a weak monoproticacid by a strong base.1V (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 thisexperiment, 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 inthe titration curve is shown on Figure 1.] The volume needed to reach the buffer point, Vbp,, willbe 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 NaOHadded. By a mathematical analysis of the titration curve, the equivalence point volume will beidentified. From this volume, the equivalent weight can be calculated. Again by mathematicalanalysis, the pH at the buffer point will be determined. This is a measure of pKa of the acid.

Question

Using the two images attached please write one paragraph explaining the analysis Also write another paragraph of the complete method in your own wordsPlease please please answer as fast as possible EXPERIMENT 8 pK AND EQUIVALENT WEIGHT BY POTENTIOMETRICTITRATION OF AN ACIDPURPOSEIn this experiment you are to determine the pK and equivalent weight of a weak monoprotic acidby potentiometric titration.METHODFor an acid, the equivalent weight is defined as the mass in grams that provides one mole ofprotons in a neutralization reaction. Thus, for a polyprotic acid, the equivalent weight variesaccording to the neutralization reaction occurring and the amount of base allowed to react withthe acid. For a monoprotic acid, HA, one mole of acid will provide one mole of protons, andtherefore the equivalent weight equals the molar mass. Note that this is only true because the acidis 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 ofacid will be reacted with a solution of the strong base, sodium hydroxide, whose molarity isknown. By determining the volume of NaOH needed to react completely with HA followingEquation (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 calledtitration. In this titration small, measured amounts of the NaOH solution are added to the solutionof HA. The point in the titration at which the moles of added base just equal the moles of addedacid 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 molarmass) of the acid.ер,Several methods can be used to detect the equivalence point. In this experiment we will followthe pH of the solution as the base is added. A typical titration curve is shown in figure 1. Noticethat at the equivalent point the pH &#34;jumps.&#34; A mathematical analysis of the data collected in thisexperiment 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. ofNaOH for the titrationof a weak monoproticacid by a strong base.1V (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 thisexperiment, 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 inthe titration curve is shown on Figure 1.] The volume needed to reach the buffer point, Vbp,, willbe 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 NaOHadded. By a mathematical analysis of the titration curve, the equivalence point volume will beidentified. From this volume, the equivalent weight can be calculated. Again by mathematicalanalysis, the pH at the buffer point will be determined. This is a measure of pKa of the acid.

Accepted Answer

The objective of this experiment is to determine the  (acid dissociation constant) and the…