Answer:
a) mol Ni(OH)2 = 3.335 E-13 mol
b) mol Cu(OH)2 = 5.497 E-8 mol
c) mol BaF2 = 2.75 E-5 mol
Step-by-step explanation:
a) Ni(OH)2 ↔ Ni2+ + 2OH-
S S 2S + [ OH- ]....................in equil. ∴ S: Solubility [=] mol/L
NaOH ↔ Na+ + OH-
∴ pH = 14 - pOH = 12.34
⇒ pOH = 14 - 12.34 = 1.66
∴ pOH = - log [ OH- ] = 1.66
⇒ [ OH- } = 0.0218 M
⇒ pKsp = - log Ksp = 15.8
⇒ Ksp = 1.585 E-16 = [ N12+ ] * [ OH- ]² = S * (2S + 0.0218)²
we assume:
if we compare the concentration (0.0218) with the value of Ksp which is in the order of E-16; we can despise the solubility as adding, obtaining:
⇒ Ksp 0 1.585 E-16 = S * ( 0.0218 )²
⇒ 1.585 E-16 = 4.752 E-4*S
⇒ S = 3.335 E-13 mol/ L * ( 1L sln) = 3.335 E-13 mol
⇒ % S = ( 3.335 E-13 / 0.0218 ) * 100 = 1.53 E-9 %....the assumption can be accepted
b) Cu(OH)2 ↔ Cu2+ + 2(OH-)
S S 2S + [ OH- ]
NaOH ↔ Na+ + OH-
∴ pH = 8.23 = 14 - pOH
⇒ pOH = 5.77
⇒ [ OH- ] = 1.698 E-6
∴ pKsp = 18.8 = - log Ksp
⇒ Ksp = 1.585 E-19 = [ Cu2+ ] * [ OH- ]² = S * ( 2S + 1.698 E-6)²
taking the same consideration as in the previous point:
⇒ 1.585 E-19 = S * ( 1.698 E-6 )²
⇒ S = 5.497 E-8 mol/L * 1 L = 5.497 E-8 mol
⇒ % S = ( 5.497 E-8 / 1.698 E-6 ) * 100 = 3.23 % < 5%; .... the assumption can be accepted
c) BaF2 ↔ Ba+ + 2F-
S S 2S + 0.12
NaF ↔ Na+ + F-
0.12 0.12 0.12
⇒ Ksp = [ Ba+ ] * [ F- ]² = S * ( 2S + 0.12 )²
∴ pKsp = - log Ksp = 5.8
∴ being the concentration >>> Ksp, as in the previous points, we depress the solubility as adding and the equation would be:
⇒ Ksp = 1.585 E-6 = S * ( 0.12 )²
⇒ Ksp = 1.585 E-6 = 0.0144*S
⇒ S = 1.100 E-4 mol/L * 0.25 L sln = 2.75 E-5 mol
⇒ % S = ( 1.10 E-4 / 0.12 ) * 100 = 0.092 %....the assumption can be accepted