The B_4O_5(OH)_4^2- ion, present in 5.0 mL of a saturated Na_2B_4O_5(OH)_4 solut
ID: 1063642 • Letter: T
Question
The B_4O_5(OH)_4^2- ion, present in 5.0 mL of a saturated Na_2B_4O_5(OH)_4 solution at a measured temperature, is titrated to the bromocresol green endpoint with 5.28 mL of 0.182 M HC1. Express all calculations with the correct number of significant figures. How many moles of B_4O_5(OH)_4^2- are present in the sample? What is the molar concentration of B_4O_5(OH)_4^2- in the sample? Calculate the K_sp for Na_2B_4O_5(OH)_4 from these data. What is the free energy change for the dissolution of Na_2B_4O_5(OH)_4 at 25 degree C? R = 8.314 times 10^-3 kJ/mol*K.Explanation / Answer
a)
moles of ion:
1 mol of ion = 4 mol of H+
mol of H+ = MV = 5.28*10^-3)(0.182) = 0.00096096 mol of ions
b)
Molar concentration
M = mol/V = 0.00096096 / (5*10^-3) = 0.192192 M
c)
Ksp = [Na+]^2 [ B4O5(OH)4)-2]
[Na+] = 2*0.192192 = 0.384384
[ B4O5(OH)4)-2] = 0.192192
Ksp =(0.384384 ^2)(0.192192 ) = 0.028396
d)
dG for...this
dG = -RT*ln(Ksp)
dG = -8.314*298*ln(0.028396) = 8823.889J /mol = 8082 kJ/mol
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