20 ml of 0.100 M weak acid is titrated with a 0.100 M NaoH solution. Here is dat

Question

20 ml of 0.100 M weak acid is titrated with a 0.100 M NaoH solution. Here is data showing the as various quantities of base are added: What is the Ka of the acid? Which of the above points is he equivalence point? Silver sulfate, Ag_2SO_4, dissolves to the extent of 4.84 grams per liter. What is the Ksp of silver sulfate from this data? 50.0 ml of a solution containing 0, 010M Pb(NO_3) is mixed with 50.0 ml of a solution containing 0.10 M NaCl. What will the concentrations of Pb^2+ and of Cl^- be after the mixing? PbCl_2 has a Ksp = 1.7 times 10^-5. What is the value of Q for the lead and chloride ions? Will a precipitate form in the mixture? Why or why not.

Explanation / Answer

46. For the titration of wek acid and strong base

A) moles of acid = 0.1 M x 20 ml = 2 mmol

at half equivalence 1 mmol of acid is neutralized

Volume of NaOH needed = 1 mmol/0.1 M = 10 ml

So at 10 ml NaOH, we have pH = pKa

pKa = 3.74

pKa = -log[Ka]

Ka = 1.82 x 10^-4

B) Equivalence point for the given titration = 20 ml NaOH added

47. Ksp of Ag2SO4 = [Ag+]^2.[SO4^2-]

with,

[Ag2SO4] in solution = 4.84 g/311.8 g/mol x 1 L

                                  = 0.0155 M

So,

Ksp = (2 x 0.0155)^2.(0.0155)

       = 1.50 x 10^-5

48. Mixing solutions

after mixing

[Pb2+] = 0.01 M x 50 ml/100 ml = 0.005 M

[Cl-] = 0.1 M x 50 ml/100 ml = 0.005 M

Qsp = [Pb2+][Cl-]^2

       = (0.005)(0.005)^2

       = 1.25 x 10^-7

Ksp for PbCl2 = 1.7 x 10^-5

So,

Qsp is lower than Ksp, thus no precipitate would form from this solution.

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