A) how many moles of ni(oh)2 are soluble in 1l of sodium hydroxide (naoh) when the ph is 12.34? pksp of ni(oh)2 = 15.8b) how many moles of cu(oh)2 are soluble in 1l of sodium hydroxide (naoh) when the ph is 8.23? pksp of cu(oh)2 = 18.8c) how many moles of baf2 are soluble in 250ml of sodium fluoride (naf)? concentration of naf = 0.12 mol/pksp of baf2 = 5.8

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

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

b) nuclear ive learned that in a pedtrodic table

the objects have the same charge is correct.

nathan lee