Please show calculations. 200.0 mL of 2.00 M solution of a weak acid is mixed wi

Question

Please show calculations.

200.0 mL of 2.00 M solution of a weak acid is mixed with 300.0 mL of a 2.00 M solution containing its conjugate base. 50.000 g of the slightly soluble salt silver acetate (AgC2H3O2) is added to this buffer solution. After thorough mixing, the solution is assumed to be saturated. To determine the amount of the solid which remains undissolved, the solution is filtered and residue is collected and dried; the mass of the undissolved solid is 30.369 g.

a. Based on this information, calculate Ksp' for silver acetate in this buffer.

b. Using your value for Ksp', determine the pH of the buffer solution.

c. Based on the pH of the buffer solution, identify the weak acid used to produce it.

Explanation / Answer

mass silver acetate dissolved part = total mass - undissolved part

                                                         = 50 -30.369

                                                        = 19.631 g

silver acetate molecular weight = 166.9 g / mol

moles of silver acetate = 19.631 / 166.9 = 0.118

molarity of silver acetate = 0.118 / volume = 0.118 / 200+300

                                        = 2.35 x 10^-4M

a)

AgC2H3O2 ---------------------> Ag+ + C2H3O2-

2.35 x 10^-4M                   2.35 x 10^-4M    2.35 x 10^-4M

Ksp = [Ag+][C2H3O2-]

Ksp = (2.35 x 10^-4)^2

Ksp = 5.53 x 10^-8

solubility product =Ksp = 5.53 x 10^-8

b)

pH = pKa + log [salt/acid]

pH = 4.74 + log [300 x 2/200 x2]

pH = 4.92

pH of buffer 4.92

C)

buffer pH is nearer to pKa of acetic acid

so acid is acetic acid

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