To solve this questions you first need to find the number of moles of barium phosphate you have. The molar mass of barium phosphate is 601.93g/mol.
24.4/601.83 = 0.0402 moles barium phosphate
Then you need to use avagadro’s number, 6.022 x 10^23, which is the number of molecules or formula units in a mole.
6.022 x 10^23 * 0.0402 = 2.42 x 10^22 formula units
Answer:
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When PH + POH = 14
∴ POH = 14 -7 = 7
when POH = -㏒[OH-]
7 = -㏒ [OH-]
∴[OH-] = 10^-7
by using ICE table:
Mn(OH)2(s) ⇄ Mn2+ (aq) + 2OH-(aq)
initial 0 10^-7
change +X +2X
Equ X (10^-7 + 2X)
when Ksp = [Mn2+][OH-]^2
when Ksp of Mn(OH)2 = 4.6 x 10^-14
by substitution:
4.6 x 10^-14 = X*(10^-7+2X)^2 by solving this equation for X
∴ X =2.3 x 10-5 M
∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol
= 0.002 g/ L
