In: Chemistry
Sodium nitrate has a solubility of 92.1 g/100 ml of water at 20°C. Calculate the solubility product.
solubility = 92.1g/100ml
= 92.1g/100*10^-3L [ 1ml = 10^-3L]
= 92.1g/0.1L
= 921g/L /85g/mole [ gram molar mass of NaNO3 = 85g/mole]
= 10.83g/L
NaNO3 ----------------> Na^+ + NO3^-
s s
Ksp = [Na^+][NO3^-]
Ksp = s*s
Ksp = 10.83*10.83 = 117.28 >>>>answer