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

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.  

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa