comibiscmsmodibisviewphpid5047041 Jun Periodic Table Questio

.com/ibiscms/mod/ibis/view.php?id-5047041 Jun Periodic Table Question 5 of 9 Map What conce solution at pH 5.0? Note that the concentration and/or pH value may difer from that in the first question earrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A W 2. Use the mole of acetate to calculate the concentration of acetate 3. Calculate the concentration of acetic acid Step 1: The ratio of base to acid is 1.7. of acetate for each molecule of Step 2: The mole fraction of acetate is 0.63, and the concentration of acetate is 0.13 That is, there are 1.7 molecules acetic acid Step 3: Calculate the concentration of acetic acid Number lacetic acid- Check Amswer Next Ex

Solution

Answer:

. Using Henderson- Hasselbalch equation-

            pH = pKa + log ([A^-] / [AH])                    - equation 1

            where, [A^-] = acetate ion

[AH] = acetic acid

Putting the values in equation 1-

            5.0 = 4.76 + log ([A^-] / [AH])

            Or, 5.0 – 4.76 = log ([A^-] / [AH])

            Or, [A^-] / [AH] = antilog 0.24 = 1.738

            Hence, [A^-] / [AH] = 1.738

# From [A^-] / [AH] = 1.738 –

            [A^-] = 1.738 [AH]

Since, all the chemical species are present in the same solution, the volume of solution remains constant. Let the volume of solution be L liters.

            Number of moles = Molarity x Volume of solution in liters

                                                = M x L

Mole fraction of A^- = Moles of [A^-] / (Moles of [A-] + Moles of [AH])

                                    = [A^-] x L / ([A-] x L + [AH] x L)

                                    = [A^-] L / { L ( [A^-] + [AH] ) }

                                    = [A^-] / ([A^-] + [AH])                      ; [we have, [AH] = 1.738 [AH]]

                                    = 1.738 [AH] / (1.738 [AH] + [AH])

                                    = 1.738 [AH] / 2.738 [AH]

                                    = 0.6347

Hence, mole fraction of Acetate ion = 0.635

# Similarly,

Mole fraction of acetic acid = [AH] / ([A^-] + [AH])

                                                            = [AH] / 2.738 [AH]

                                                            = 0.365

Hence, mole fraction of acetic acid = 0.365

# [Acetic acid] = Molarity of buffer x mole fraction of acetic acid

                        = 0.20 M x 0.365

                        = 0.073 M