4. 5 points) 25.00 mL of a solution of 0.25 M HA is titrated with 0.25 M NaOH di

Question

4. 5 points) 25.00 mL of a solution of 0.25 M HA is titrated with 0.25 M NaOH dispensed from a buret. The acid HA is a weak acid with pK,-4.92. Which one of the following indicators would be the best choice for detection of the equivalence point in this titration? Color In Acid Red Yellow Orange Red Yellow pH Range 1.2-2.8 3.0-4.6 3.1-4.4 4.2-6.3 4.8-6.4 6.0-7.6 7.2-8.8 8.3-10.0 In Base Yellow Bluish purple Thymol blue Bromophenol blue Methyl orange Methyl red Chlorophenol blue Bromothymol blue Cresol red Yellow Red Blue Red Reddish pink 5. (10 points) The Ksp of PbF2 is 4.1 x 10 A. Write a reaction representing the dissolution of PbF2 (s) B. Write the Ksp equilibrium constant expression C. What is IPb2+1 in a saturated solution of PbF2 (s) in pure water? D. What is (Pb in a saturated solution of PbF2 (s) in 0.05 M NaF (aq)?

Explanation / Answer

a)

Solubility is nothing more than the solubiltiy into auqeous ions; identify ions (cation/anion)

PbF2(s) <--< Pb2+(aq) + 2F-(aq)

b)

Ksp expression is nothing more than the ions related ot Ksp

Ksp = [PB2+][F-]^2

c)

if pure water, then

1 mol of PBF2 = 1 mol of Pb2+

1 mol of PBF2 = 2 mol of F-

Ksp = (S)(2S)^2

4*S^3 = Ksp

S = (Ksp/4)^(1/3)

S = ((4.1*10^-8)/4)^(1/3)

S = 0.002172 M of PbF2

d)

if [F-] = [naF] = 0.05 M

then

Ksp = [Pb2+][F-]^2

4.1*10^-8 = (S)(0.05)^2

S = (4.1*10^-8) / ((0.05)^2) = 0.0000164 M of PbF2

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