Question

17.29 The solubility of silver bromate, AgBrO3, in water is 0.0072 g/L. Calculate Ksp. (MW of AgBroz is 236 g/mol). Hint - fi
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Answer #1

Molar mass of AgBrO3 = 236 g/mol

Molar mass of AgBrO3= 235.8 g/mol

s = 7.2*10^-3 g/L

To covert it to mol/L, divide it by molar mass

s = 7.2*10^-3 g/L / 236 g/mol

s = 3.05*10^-5 mol/L

At equilibrium:

AgBrO3 <----> Ag+ + BrO3-

   s s

Ksp = [Ag+][BrO3-]

Ksp = (s)*(s)

Ksp = 1(s)^2

Ksp = 1(3.05*10^-5)^2

Ksp = 9.3*10^-10

Answer: 9.3*10^-10

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