7.67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

1. Silver chromate:Ag2CrO4→2Ag++CrO42−Then, [Ag+] = (2s), [CrO42−] = sKsp= [Ag+]2 [CrO42−]s = 0.65×10−4MSo, [Ag+] = 2s = 1.30 x 10−4M, [CrO42−] = 6.5 x 10-5M2. Barium Chromate:BaCrO4→Ba2++CrO42− [Ba2+] = [CrO42−] = sThen, Ksp= [Ba2+] [CrO42−] = s x s= > 1.2 x 10-10M = s2= > s = 1.09 x 10-5M3. Ferric Hydroxide:Fe (OH)3→Fe3+ + 3OH−Then [Fe3+] = s, [OH−] = 3sKsp= [Fe3+] [OH−]3Let s be the solubility of Fe (OH)3 [Fe3+] = s = 1.38 x 10-10M [OH−] = 3s = 4.14 x 10-10M4. Lead Chloride:PbCl2→Pb2++2Cl−Then, [Pb2+] = s, [Cl−] = 2sKsp= [Pb2+] [Cl−]2= >Ksp=s x (2s)2 =4s3 ⇒1.6×10−5=4s3⇒0.4×10−5=s3 =>4×10−6=s3 [Pb2+] = s=1.58×10−2M [Cl−] = 2s=3.16×10−2M5. Mercurous Iodide:Hg2I2→ 2Hg++2I− [Hg2+] = [I−] = 2sKsp= [Hg2+]2 [I−]2=>4.5 x 10-29 = (2s)2 (2s)2 = 16s4 => s = 4.09 x 10-8 M Therefore, [Hg2+] = [I−] = 2s = 8.18 x 10-8 M

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Ncert Solutions Chemistry Class 11th Chemistry Class 11th Equilibrium

7.67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

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Vishal Baghel | Contributor-Level 10

4 months ago

1. Silver chromate:

Ag₂CrO₄→2Ag⁺+CrO₄²⁻

Then, [Ag⁺] = (2s), [CrO₄²⁻] = s

K_sp= [Ag⁺]² [CrO₄²⁻]

s = 0.65×10⁻⁴M

So, [Ag⁺] = 2s = 1.30 x 10⁻⁴M, [CrO₄²⁻] = 6.5 x 10^-5M

2. Barium Chromate:

BaCrO₄→Ba²⁺+CrO₄²⁻

[Ba²⁺] = [CrO₄²⁻] = s

Then, K_sp= [Ba²⁺] [CrO₄²⁻] = s x s

= > 1.2 x 10^-10M = s²

= > s = 1.09 x 10^-5M

3. Ferric Hydroxide:

Fe (OH)₃→Fe³⁺ + 3OH⁻

Then [Fe³⁺] = s, [OH⁻] = 3s

K_sp= [Fe³⁺] [OH⁻]³

Let s be the solubility of Fe (OH)₃

[Fe³⁺] = s = 1.38 x 10^-10M

[OH⁻] = 3s = 4.14 x 10^-10M

4. Lead Chloride:

PbCl₂→Pb²⁺+2Cl⁻

Then, [Pb²⁺] = s, [Cl⁻] = 2s

K_s

...more

1. Silver chromate:Ag2CrO4→2Ag++CrO42−Then, [Ag+] = (2s), [CrO42−] = sKsp= [Ag+]2 [CrO42−]s = 0.65×10−4MSo, [Ag+] = 2s = 1.30 x 10−4M, [CrO42−] = 6.5 x 10-5M2. Barium Chromate:BaCrO4→Ba2++CrO42− [Ba2+] = [CrO42−] = sThen, Ksp= [Ba2+] [CrO42−] = s x s= > 1.2 x 10-10M = s2= > s = 1.09 x 10-5M3. Ferric Hydroxide:Fe (OH)3→Fe3+ + 3OH−Then [Fe3+] = s, [OH−] = 3sKsp= [Fe3+] [OH−]3Let s be the solubility of Fe (OH)3 [Fe3+] = s = 1.38 x 10-10M [OH−] = 3s = 4.14 x 10-10M4. Lead Chloride:PbCl2→Pb2++2Cl−Then, [Pb2+] = s, [Cl−] = 2sKsp= [Pb2+] [Cl−]2= >Ksp=s x (2s)2 =4s3 ⇒1.6×10−5=4s3⇒0.4×10−5=s3 =>4×10−6=s3 [Pb2+] = s=1.58×10−2M [Cl−] = 2s=3.16×10−2M5. Mercurous Iodide:Hg2I2→ 2Hg++2I− [Hg2+] = [I−] = 2sKsp= [Hg2+]2 [I−]2=>4.5 x 10-29 = (2s)2 (2s)2 = 16s4 => s = 4.09 x 10-8 M Therefore, [Hg2+] = [I−] = 2s = 8.18 x 10-8 M

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7.67. Determine the solubility of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

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