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Question: For a saturated solution of AgCl at 25°C, specific conductance is 3.4 x $10^{-x}$ $ohm^{-1}$ $cm^{-1...

For a saturated solution of AgCl at 25°C, specific conductance is 3.4 x 10−x10^{-x} ohm−1ohm^{-1} cm−1cm^{-1} and that of water used for preparing the solution is 1.6 x 10−610^{-6} ohm−1ohm^{-1} cm−1cm^{-1}. If ∧AgCl∞\land_{AgCl}^{\infty} = 138.3 ohm−1ohm^{-1} cm2cm^{2} eq−1eq^{-1}, the KspK_{sp} of AgCl is 0.17 x 10−910^{-9} M2M^{2}. The value of x is [Given: 1.7\sqrt{1.7} = 1.3]

Answer

6

Explanation

Solution

To solve this problem, we need to relate the given specific conductances and limiting molar conductivity to the solubility product constant (KspK_{sp}) of AgCl.

1. Calculate the specific conductance of AgCl (κAgCl\kappa_{AgCl}):

The specific conductance of the saturated AgCl solution includes the contribution from water. Therefore, to find the specific conductance due to AgCl ions, we subtract the specific conductance of water from that of the solution.

κAgCl=κsolution−κwater\kappa_{AgCl} = \kappa_{solution} - \kappa_{water}

Given:

κsolution=3.4×10−x ohm−1 cm−1\kappa_{solution} = 3.4 \times 10^{-x} \text{ ohm}^{-1} \text{ cm}^{-1}

κwater=1.6×10−6 ohm−1 cm−1\kappa_{water} = 1.6 \times 10^{-6} \text{ ohm}^{-1} \text{ cm}^{-1}

So, κAgCl=(3.4×10−x−1.6×10−6) ohm−1 cm−1\kappa_{AgCl} = (3.4 \times 10^{-x} - 1.6 \times 10^{-6}) \text{ ohm}^{-1} \text{ cm}^{-1}

2. Calculate the solubility (s) of AgCl from its KspK_{sp}:

For a sparingly soluble salt like AgCl, the dissolution equilibrium is:

AgCl(s)⇌Ag+(aq)+Cl−(aq)AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)

If 's' is the solubility of AgCl in mol/L, then [Ag+]=s[Ag^+] = s and [Cl−]=s[Cl^-] = s.

The solubility product constant is given by:

Ksp=[Ag+][Cl−]=s2K_{sp} = [Ag^+][Cl^-] = s^2

Given Ksp=0.17×10−9M2K_{sp} = 0.17 \times 10^{-9} M^2.

s2=0.17×10−9s^2 = 0.17 \times 10^{-9}

To make the exponent even for square root:

s2=1.7×10−1×10−9=1.7×10−10s^2 = 1.7 \times 10^{-1} \times 10^{-9} = 1.7 \times 10^{-10}

s=1.7×10−10s = \sqrt{1.7 \times 10^{-10}}

s=1.7×10−10s = \sqrt{1.7} \times \sqrt{10^{-10}}

Given 1.7=1.3\sqrt{1.7} = 1.3.

s=1.3×10−5 mol/Ls = 1.3 \times 10^{-5} \text{ mol/L}

3. Relate solubility (s) to specific conductance and limiting molar conductivity:

The molar conductivity (Λm\Lambda_m) is related to specific conductance (κ\kappa) and concentration (C) by the formula:

Λm=κ×1000C\Lambda_m = \frac{\kappa \times 1000}{C}

For a sparingly soluble salt, the concentration of the saturated solution is its solubility 's'. Also, the molar conductivity of a very dilute solution (like a saturated solution of a sparingly soluble salt) can be approximated by its limiting molar conductivity (Λm∞\Lambda_m^\infty).

So, s=C=κAgCl×1000ΛAgCl∞s = C = \frac{\kappa_{AgCl} \times 1000}{\Lambda_{AgCl}^\infty}

Given ΛAgCl∞=138.3 ohm−1 cm2 eq−1\Lambda_{AgCl}^\infty = 138.3 \text{ ohm}^{-1} \text{ cm}^2 \text{ eq}^{-1}. Since AgCl is a 1:1 electrolyte, eq−1eq^{-1} is equivalent to mol−1mol^{-1}.

So, ΛAgCl∞=138.3 ohm−1 cm2 mol−1\Lambda_{AgCl}^\infty = 138.3 \text{ ohm}^{-1} \text{ cm}^2 \text{ mol}^{-1}.

Substitute the values:

1.3×10−5=(3.4×10−x−1.6×10−6)×1000138.31.3 \times 10^{-5} = \frac{(3.4 \times 10^{-x} - 1.6 \times 10^{-6}) \times 1000}{138.3}

4. Solve for x:

Rearrange the equation to solve for the term containing x:

(1.3×10−5)×138.3=(3.4×10−x−1.6×10−6)×1000(1.3 \times 10^{-5}) \times 138.3 = (3.4 \times 10^{-x} - 1.6 \times 10^{-6}) \times 1000

Divide both sides by 1000:

(1.3×10−5)×138.31000=3.4×10−x−1.6×10−6\frac{(1.3 \times 10^{-5}) \times 138.3}{1000} = 3.4 \times 10^{-x} - 1.6 \times 10^{-6}

(1.3×10−5)×0.1383=3.4×10−x−1.6×10−6(1.3 \times 10^{-5}) \times 0.1383 = 3.4 \times 10^{-x} - 1.6 \times 10^{-6}

0.17979×10−5=3.4×10−x−1.6×10−60.17979 \times 10^{-5} = 3.4 \times 10^{-x} - 1.6 \times 10^{-6}

1.7979×10−6=3.4×10−x−1.6×10−61.7979 \times 10^{-6} = 3.4 \times 10^{-x} - 1.6 \times 10^{-6}

Now, isolate 3.4×10−x3.4 \times 10^{-x}:

3.4×10−x=1.7979×10−6+1.6×10−63.4 \times 10^{-x} = 1.7979 \times 10^{-6} + 1.6 \times 10^{-6}

3.4×10−x=(1.7979+1.6)×10−63.4 \times 10^{-x} = (1.7979 + 1.6) \times 10^{-6}

3.4×10−x=3.3979×10−63.4 \times 10^{-x} = 3.3979 \times 10^{-6}

Since the problem expects an integer value for x and the values are very close, we can approximate 3.3979≈3.43.3979 \approx 3.4.

3.4×10−x=3.4×10−63.4 \times 10^{-x} = 3.4 \times 10^{-6}

10−x=10−610^{-x} = 10^{-6}

Therefore, x=6x = 6.