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CaSO₄ (s)↔Ca²⁺ (aq)+SO²⁻₄ (aq)

Ksp= [Ca²⁺] [ SO²⁻₄]

Let the solubility of CaSO4 be s.

[Ca²⁺] = [ SO²⁻₄] = s

Then, Ksp=s²

9.1×10−6=s²

s =3.02×10⁻³mol/L

Molecular mass of CaSO₄=136g/mol

Solubility of CaSO₄ in gram/L= 3.02×10⁻³×136=0.41g/L

This means that we need 1L of water to dissolve 0.41g of CaSO₄.

Therefore, to dissolve 1g of CaSO₄ we require = 10.41L= 2.44Lof water.

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Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., x/2.

∴ [FeSO₄]= [Na₂S]=x / 2

Then, [Fe2⁺]= [FeSO₄]=x/ 2

Also, [S²⁻]= [Na₂S]=x/2

FeS (x)↔Fe²⁺ (aq)+S²⁻ (aq)

K_sp= [Fe²⁺] [S²⁻]

= >6.3×10⁻¹⁸= (x/2) (x/2)

x²/4=6.3×10⁻¹⁸

⇒x= 5.02×10⁻⁹

If the concentrations of both solutions are equal to or less than 5.02×10⁻⁹M, then there will be no precipitation of iron sulphide.

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After mixing, the concentration of NaIO₃?  is 0.002 / 2? =0.001M.
After mixing, the concentration of Cu (ClO_3? )_2?  is 0.002? / 2 =0.001M.
NaIO₃? Na⁺+IO₃⁻?
[IO₃⁻? ]=0.001M
Cu (ClO₃? )₂? Cu²⁺+2ClO₃⁻?
[Cu²⁺]=0.001M
The ionic product of Cu (IO₃? )₂?  is
[Cu²⁺] [I⁻]²=0.001×0.001²=1×10⁻⁹
As the ionic product is less than the solubility product, no precipitation will occur.

Silver chromate: Ag₂CrO₄? ? 2Ag⁺ + CrO₄^2−?
[Ag⁺] = 2s₁? , CrO₄²⁻? = s₁?
K_sp? = (2s₁? )² (s₁? ) = 4s³ = 1.1×10⁻¹²
s_1? = 6.5 × 10−5 . (1)
Silver bromide: AgBr? Ag⁺ + Br⁻
[Ag⁺] = [Br⁻] = s₂?
 Ksp? = (s₂? ) × (s₂? ) = (s₂)²? = 5.0 × 10⁻¹³
s_2? =7.07×10⁻⁷. (2)
Divide equation (1) by equation (2) to obtain the ratio of the molarities of saturated solutions:
? s₁/s₂? = (6.50×10⁻⁵)/ (7.07×10⁻⁷)? = 91.9 

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Ionic product,

K_w= [H⁺] [OH⁻]

Let [H⁺]= x

Since  [H⁺]= [OH⁻], Kw=x². 

⇒K_w at 310K is 2.7×10⁻¹⁴

∴2.7×10⁻¹⁴=x²

⇒x=1.64×10⁻⁷

⇒ [H⁺]=1.64×10⁻⁷

⇒ pH= −log [H⁺]=−log [1.64×10⁻⁷]=6.78

Hence, the pH of neutral water is 6.78.

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Given K_a=1.35 x 10⁻³

For acid solution:

[H⁺] = (K_aC)^1/2 = (0.00135 x 0.1)^1/2 = 0.0116M

   pH = – log [H⁺] = –log0.0116 = 1.936

For 0.1M sodium salt solution

ClCH₂COONa is the salt of a weak acid i.e., ClCH₂COOH and a strong base i.e., NaOH.

pK_a = -logK_a = log (0.00135) = 2.8697

pK_w = 14

logc = log0.1 = −1

pH = 0.5 [pK_w + pK_a + logc] = 0.5 [14 + 2.8697 -1]

= 7.935

(i) NaCl:

NaCl+H₂O↔NaOH+HCl

Strong base Strong acid

Therefore, it is a neutral solution.

(ii) KBr:

KBr+H₂O↔KOH+HBr

Strong base  Strong acid

Therefore, it is a neutral solution.

(iii) NaCN:

NaCN+H₂O↔HCN+NaOH

Weak acid  Strong base

Therefore, it is a basic solution.

(iv) NH₄NO₃

NH₄NO₃+H₂O↔NH₄OH+HNO₃

Weak base    Strong acid

Therefore, it is an acidic solution.

(v) NaNO₂

NaNO₂+H₂O↔NaOH+HNO₂

Strong base    Weak acid

Therefore, it is a basic solution.

(vi) KF

KF+H₂O↔KOH+HF

Strong base   Weak acid

Therefore, it is a basic solution.

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