Hoping you could help me with a chemistry problem. I have a quiz tomorrow and am

Question

Hoping you could help me with a chemistry problem. I have a quiz tomorrow and am unsure on how to solve this problem. The B4O5(OH)4 ^2- ion, present in 5.0 mL of a saturated Na2B4O5(OH)4 solution at a measured temperature, is titrated to the bromocresol green endpoint with 5.28 mL of .182 M HCl. Express all calculations with the correct number of significant figures. a) How many moles of B4O5(OH)4 ^2- are present in the sample? b) What is the molar concentration of B4O5(OH)4 ^2- in the sample? c) Calculate Ksp for Na2B4O5(OH)4 from these data. d) What is the free energy change for the dissolution of Na2B4O5(OH)4 at 25 degrees celcius? R= 8.314 x 10^-3 kj/mol K

Explanation / Answer

For, Na2B4O5(OH)4 ionization reaction can be written as,

Na2B4O5(OH)4 ----------- > 2Na+ + B4O5(OH)42-

Being a strong electrolyte we have,

[B4O5(OH)42-] = [Na2B4O5(OH)4]

[Na+] = 2 x [Na2B4O5(OH)4] = 2 x [B4O5(OH)42-]……….(1)

1) [B4O5(OH)42-] = ?

For [B4O5(OH)42-] ; V1 = 5.0 mL, M1 = ? and milimoles of [B4O5(OH)42-] = M1V1

For HCl ; V2 =5.28 mL, M2 = 0.182 M. and Milimoles of HCl = M2V2

At end point,

[B4O5(OH)42-] = [HCl]

i.e. Milimoles of B4O5(OH)42- = Milimoles of HCl

i.e. M1V1 = M2V2

so, M1 x 5.0 = 0.182 x 5.28

M1 x 5.0 = 0.96096

M1 = 0.96096 / 5.0

M1 = 0.1922 M

Hence 0.1922 moles of [B4O5(OH)42-] present in the sample.

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2) Molar concentration B4O5(OH)42- =?

Molar concentrations are expressed in M/L

We have, 0.1922 M In 5mL i.e. concentration is 0.1922 M/ 5mL, hence

Molar concentration = 0.1922 x 200 / 5mL x 200 ………………….(multiplied by 200 to make volume 1000 as 200 x 5 = 1000)

Molar concentration = 38.44 M/ 1000 mL

Molar concentration of B4O5(OH)42- = 38.44 M/L

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3)[Na2B4O5(OH)4] ; Ksp =?

Na2B4O5(OH)4, Ionizes according to the equation,

Na2B4O5(OH)4 ----------- > 2Na+ + B4O5(OH)42-

Ksp = [B4O5(OH)42-] x [Na+]2

We have [Na+] = 2 x [B4O5(OH)42- ]

Hence we write,

Ksp = [B4O5(OH)42-] x (2 x [B4O5(OH)42- ])2

Ksp = 4 x ([B4O5(OH)42-])3

Ksp = 4 x (38.44)3 (M/L)3

Ksp =2.27 x 105 (M/L)3

Solubility product value for Na2B4O5(OH)4 is 2.27 x 105 (M/L)3

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4)Free energy change i.e. Gibbs free energy change for dissolution of Na2B4O5(OH)4 is related to Ksp as,

G0 = -RTln(Ksp)

We have, T = 25 0C = 25 + 273.15 = 298.15 K

R = 8.314 x 10-3 kJ.K-1.mol-1.

Ksp = 2.27 x 105

Let us put all these values in above equation,

G0 = -8.314 x 10-3 x 298.15 x ln(2.27 x 105)

G0 = -30.57 kJ……………..(as we used R value in kJ.K-1.mol-1 unit we get energy in kJ directly)

Hence standard free energy change for dissolution of Na2B4O5(OH)4 is -30.57 kJ.

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