Assume no volume change, and calculate the following for 75 mL of 1.4 M Ba(NO3)3

Question

Assume no volume change, and calculate the following for 75 mL of 1.4 M Ba(NO3)3 assume KSP (BaF2) equals 1.7×10^-6 A) how many milligrams of NaF must be added to begin the precipitation? B) how many grams of NaF must be added to precipitate 17.5g of BaF2? C) how many grams of NaF must be added to reduce the Ba(NO3)2 concentration to .76 M? Assume no volume change, and calculate the following for 75 mL of 1.4 M Ba(NO3)3 assume KSP (BaF2) equals 1.7×10^-6 A) how many milligrams of NaF must be added to begin the precipitation? B) how many grams of NaF must be added to precipitate 17.5g of BaF2? C) how many grams of NaF must be added to reduce the Ba(NO3)2 concentration to .76 M? Assume no volume change, and calculate the following for 75 mL of 1.4 M Ba(NO3)3 assume KSP (BaF2) equals 1.7×10^-6 A) how many milligrams of NaF must be added to begin the precipitation? B) how many grams of NaF must be added to precipitate 17.5g of BaF2? C) how many grams of NaF must be added to reduce the Ba(NO3)2 concentration to .76 M?

Explanation / Answer

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

so

a)

Find F- so precipitations starts, so this is at equilibrium:

1.7*10^-6 = (1.4)(F-)^2

[F-] = sqrt((1.7*10^-6) / 1.4 ) = 0.001101 M required

so

V = 75 mL = 75*10^-3 L

mol = MV = (0.001101)(75*10^-3) = 0.000082575 mol of F- required

mol of NAF = 0.00008257mol

mass NaF = mol*MW = (0.000082575)(41.98817) = 0.00346 g of NaF

b)

in order to precipitate 17.5 g of BaF2

mol of BaF2 = mass/MW = 17.5/(175.34 ) = 0.099806 mol of BaF2

so

mol of Ba+2 = 0.099806 mol

mol of F- = 0.099806*2 = 0.199612 mol

mol of NaF = 0.199612

mass of NaF = 41.98817*0.199612 = 8.381 g of NaF

c)

mas sof NaF to Ba+2 --> 0.76

so:

M = 0.76 M left...

mol = MV = (0.76)(75*10^-3) = 0.057 mol of Ba+2 left

mol initially Ba+2 = 1.4*75/1000 = 0.105 mol of Ba+2

change = (0.105-0.057) = 0.048 mol of Ba+e must precipitate

so:

2 mol of NaF --> 1 mol of Ba+2

mol of NaF --> 0.048*2 = 0.096 mol of NaF required

mass = mol*MW = 0.096*41.98817 = 4.0308 g of NAF

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