Question

Part A Acetic acid has a Ka of 1.8×10−5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: [acetic acid] ten times greater than [acetate], [acetate] ten times greater than [acetic acid], and [acetate]=[acetic acid]. Match each buffer to the expected pH. Drag each item to the appropriate bin. Part B How many grams of dry NH4Cl need to be added to 2.40 L of a 0.100 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.78? Kb for ammonia is 1.8×10−5.

Answer 1

Acetic acid has K 1.8x103 Three buffer solutions made with varying concentrations then, calculate pH of each buffer as follows Buffer A: [acetic acid] ten times greater than [acetate] Now, calculate the pH of the buffer A solution from the following Henderson Hasselbalch equation. base pH-pK, +log acid First calculate the value of pKa of acetic acid from the following equation. log(1.8x10-3 --(-4.744) 4.74 Then, the value pKa of acetic acid is 4.74 Now, substitute 4.74 for pKa in equation1 base lacid) = 4.74 + log acetate aceti cacid acetat = 4.74 + log 10 acetate Thus, the pH of the buffer A solution is 3.74

Buffer B: [acetate] ten times greater than [acetic acid] Now, calculate the pH of the buffer B solution, substitute 4.74 for pK in equation 1 pH-pK, +10g-ase [acid] acetate = 4.74 + log aceticaad 10 acetic acid - 4.74 log aceticacid 5.74 Thus, the pH of the buffer B solution is 5.74 Buffer C: [acetatel [acetic acid] Now, calculate the pH of the buffer C solution, substitute 4.74 for pK in equation 1 pH-pKa+log base 4[acid - 4.74+log -4.74+log - 4.74 acetate aceti cacid acetic acid aceti cacid Thus, the pH of the buffer B solution is 4.74 Part B The value of K, of ammonia is K,-1.8x10- Calculate the mass of dry NH4C1 added to 2.400 L of a 0.100 M of ammonia with pH 8.78

Ammonia is base and NH4Cl is salt. So, calculate the pOH of ammonia from the value of pH as follows pH+pOH 14 pOH 14-8.78 -5.22 Thus, the value of pOH is 5.22. Now, calculate the ratio of concentration of NH4C to the NH from the following Henderson-Hasselbalch equation. [salt] pOH-pK, +log NH4C 5.22log 1.8x10og log522-4.74 NH - 0.48 Then, NH4Cl NH lo 0.48 [NH,CI]10 NH -3.019 Now, calculate the molarity of NH,C1 as follows NH,CI NH - 3.019 [NH C]-3.019x0.100 M 0.3019 M -0.302 M Thus, the molarity of NH4C1 0.302 M

Now, calculate the number of moles of NH.C1 as follows mol Moles of NH C 0.302x2.40 L 0.7245 mol 0.725 mol Then, comvet moles to gams from the following equation. Mass of NH4Cl in grams Number of molesx Molecular weight -0.725 mol x53491 g/mol -39.78 g Thus, 39.78 grams of NH4Cl is needed

Answer 2

PKa = -logKa

         = -log1.8*10^-5

         = 4.7447

PH   = Pka + log[acetate]/[acetic acid]

          = 4.7447+ log1/10

         = 4.7447-1 = 3.7447

part-B

   no of moles of NH3 = molarity* volume in L

                                    = 0.1*2.4 = 0.24 moles

                         Pkb    = -log1.8*10^-5

                                    = 4.7447

                         PoH = 14-PH

                                = 14-8.78 = 5.22

                  POH   = Pkb + log[NH4Cl]/[NH3]

                 5.22    = 4.7447 + log[NH4Cl]/0.24

   5.22-4.7447 = log[NH4Cl]/0.24

                log[NH4Cl]/0.24 = 0.4753

                [NH4Cl]/0.24         = 10^0.4753

                 [NH4Cl]/0.24   = 2.9874

                [NH4Cl ]         = 2.9874*0.24 = 0.717moles

              mass of NH4Cl = no of moles* gram molar mass

                                        = 0.717*53.5 = 38.36g