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Question: The plot of $\frac{1}{\Lambda_m}$ against $c\Lambda_m$ for aqueous solution of weak monobasic acid (...

The plot of 1Λm\frac{1}{\Lambda_m} against cΛmc\Lambda_m for aqueous solution of weak monobasic acid (HX) resulted in following graph. If Λm∘\Lambda_m^\circ for HX = 400 S cm2^2 mol−1^{-1}. The degree of dissociation when c = 0.1 M is a ×\times 10−3^{-3}. Value of a is _______. (Nearest integer)

Answer

14

Explanation

Solution

The relationship between molar conductivity (Λm\Lambda_m), limiting molar conductivity (Λm∘\Lambda_m^\circ), concentration (cc), and degree of dissociation (α\alpha) for a weak electrolyte is given by Ostwald's Dilution Law.

  1. Ostwald's Dilution Law:
    For a weak monobasic acid HX, the dissociation equilibrium is:
    HX⇌H++X−HX \rightleftharpoons H^+ + X^-
    The dissociation constant KaK_a is given by:
    Ka=[H+][X−][HX]=cα21−αK_a = \frac{[H^+][X^-]}{[HX]} = \frac{c\alpha^2}{1-\alpha}

  2. Relation between α\alpha and Λm\Lambda_m:
    The degree of dissociation α\alpha is also related to molar conductivities:
    α=ΛmΛm∘\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

  3. Deriving the linear relationship:
    Substitute α\alpha from the second equation into the first equation:
    Ka=c(ΛmΛm∘)21−ΛmΛm∘K_a = \frac{c \left(\frac{\Lambda_m}{\Lambda_m^\circ}\right)^2}{1 - \frac{\Lambda_m}{\Lambda_m^\circ}}
    Ka=cΛm2(Λm∘)2(Λm∘−ΛmΛm∘)K_a = \frac{c \Lambda_m^2}{(\Lambda_m^\circ)^2 \left(\frac{\Lambda_m^\circ - \Lambda_m}{\Lambda_m^\circ}\right)}
    Ka=cΛm2Λm∘(Λm∘−Λm)K_a = \frac{c \Lambda_m^2}{\Lambda_m^\circ (\Lambda_m^\circ - \Lambda_m)}

    Rearrange the equation to match the given plot of 1Λm\frac{1}{\Lambda_m} vs cΛmc\Lambda_m:
    KaΛm∘(Λm∘−Λm)=cΛm2K_a \Lambda_m^\circ (\Lambda_m^\circ - \Lambda_m) = c\Lambda_m^2
    Divide both sides by Λm\Lambda_m:
    KaΛm∘(Λm∘Λm−1)=cΛmK_a \Lambda_m^\circ \left(\frac{\Lambda_m^\circ}{\Lambda_m} - 1\right) = c\Lambda_m
    Divide both sides by KaΛm∘K_a \Lambda_m^\circ:
    Λm∘Λm−1=cΛmKaΛm∘\frac{\Lambda_m^\circ}{\Lambda_m} - 1 = \frac{c\Lambda_m}{K_a \Lambda_m^\circ}
    Λm∘Λm=1+cΛmKaΛm∘\frac{\Lambda_m^\circ}{\Lambda_m} = 1 + \frac{c\Lambda_m}{K_a \Lambda_m^\circ}
    Divide both sides by Λm∘\Lambda_m^\circ:
    1Λm=1Λm∘+cΛmKa(Λm∘)2\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^\circ} + \frac{c\Lambda_m}{K_a (\Lambda_m^\circ)^2}

    This equation is in the form y=mx+by = mx + b, where:
    y=1Λmy = \frac{1}{\Lambda_m}
    x=cΛmx = c\Lambda_m
    Slope (mm) =1Ka(Λm∘)2= \frac{1}{K_a (\Lambda_m^\circ)^2}
    Y-intercept (bb) =1Λm∘= \frac{1}{\Lambda_m^\circ}

  4. Calculate KaK_a:
    Given:
    Slope =516= \frac{5}{16} mol L−1^{-1}
    Λm∘=400\Lambda_m^\circ = 400 S cm2^2 mol−1^{-1}

    We need to ensure consistent units. The slope is given in mol L−1^{-1}, which is equivalent to mol dm−3^{-3}.
    Λm∘\Lambda_m^\circ is in S cm2^2 mol−1^{-1}. To make units consistent, we can convert Λm∘\Lambda_m^\circ to S dm2^2 mol−1^{-1} or convert the slope.
    Let's use Λm∘\Lambda_m^\circ in S cm2^2 mol−1^{-1} and ensure KaK_a units are consistent.
    KaK_a typically has units of mol L−1^{-1} (or M).
    The term Ka(Λm∘)2K_a (\Lambda_m^\circ)^2 will have units of (mol L−1^{-1}) (S cm2^2 mol−1^{-1})2^2 = mol L−1^{-1} S2^2 cm4^4 mol−2^{-2} = S2^2 cm4^4 L−1^{-1} mol−1^{-1}.
    Therefore, the slope 1Ka(Λm∘)2\frac{1}{K_a (\Lambda_m^\circ)^2} should have units of S−2^{-2} cm−4^{-4} L mol.
    However, the given slope is in mol L−1^{-1}. This implies that the equation used in the graph implicitly assumes a certain unit conversion factor or that the units for the slope are specific to the graph's context.
    Let's assume the given slope value directly corresponds to 1Ka(Λm∘)2\frac{1}{K_a (\Lambda_m^\circ)^2} without further unit conversions for the numerical value, or that the KaK_a value will be obtained in units consistent with the slope.
    The most common form of Ostwald's dilution law for KaK_a is Ka=cα21−αK_a = \frac{c\alpha^2}{1-\alpha}, where cc is in mol/L and KaK_a is in mol/L.
    The equation 1Λm=1Λm∘+cΛmKa(Λm∘)2\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^\circ} + \frac{c\Lambda_m}{K_a (\Lambda_m^\circ)^2} is often used with Λm\Lambda_m in S cm2^2 mol−1^{-1}, cc in mol L−1^{-1} and KaK_a in mol L−1^{-1}.
    In this case, the units of 1Ka(Λm∘)2\frac{1}{K_a (\Lambda_m^\circ)^2} would be 1(mol L−1)(S cm2mol−1)2=Lmol S2cm4\frac{1}{(\text{mol L}^{-1}) (\text{S cm}^2 \text{mol}^{-1})^2} = \frac{\text{L}}{\text{mol S}^2 \text{cm}^4}. This is not mol L−1^{-1}.

    Let's re-examine the units of the slope given. If the slope is 1/(Ka(Λm∘)2)1/ (K_a (\Lambda_m^\circ)^2), and the units of KaK_a are mol/L, and Λm∘\Lambda_m^\circ are S cm2^2 mol−1^{-1}, the units of the slope should be Lmol S2cm4\frac{\text{L}}{\text{mol S}^2 \text{cm}^4}.
    However, if cc is expressed in M (mol/L) and Λm\Lambda_m in S cm2^2 mol−1^{-1}, then cΛmc\Lambda_m has units of M S cm2^2 mol−1^{-1} = (mol/L) S cm2^2 mol−1^{-1} = S cm2^2 L−1^{-1}.
    And 1/Λm1/\Lambda_m has units of S−1^{-1} cm−2^{-2} mol.
    So, the slope Δ(1/Λm)Δ(cΛm)\frac{\Delta(1/\Lambda_m)}{\Delta(c\Lambda_m)} would have units of S−1cm−2molS cm2L−1=S−2cm−4mol L\frac{\text{S}^{-1} \text{cm}^{-2} \text{mol}}{\text{S cm}^2 \text{L}^{-1}} = \text{S}^{-2} \text{cm}^{-4} \text{mol L}.
    The given slope is 516\frac{5}{16} mol L−1^{-1}. This unit implies that some conversion factor for volume (cm3^3 to L) is absorbed or the units are simplified for KaK_a.

    Let's assume the given slope value is numerically correct for the expression 1Ka(Λm∘)2\frac{1}{K_a (\Lambda_m^\circ)^2} and we need to determine KaK_a.
    Slope =1Ka(Λm∘)2= \frac{1}{K_a (\Lambda_m^\circ)^2}
    516=1Ka(400)2\frac{5}{16} = \frac{1}{K_a (400)^2}
    Ka=165×(400)2=165×160000=16800000=150000K_a = \frac{16}{5 \times (400)^2} = \frac{16}{5 \times 160000} = \frac{16}{800000} = \frac{1}{50000}
    Ka=2×10−5K_a = 2 \times 10^{-5} mol L−1^{-1} (or M)

  5. Calculate degree of dissociation (α\alpha) when c=0.1c = 0.1 M:
    We use the formula Ka=cα21−αK_a = \frac{c\alpha^2}{1-\alpha}.
    Given c=0.1c = 0.1 M and Ka=2×10−5K_a = 2 \times 10^{-5} M.
    Since the acid is weak, α\alpha is usually very small, so we can approximate 1−α≈11-\alpha \approx 1.
    Ka≈cα2K_a \approx c\alpha^2
    2×10−5=0.1×α22 \times 10^{-5} = 0.1 \times \alpha^2
    α2=2×10−50.1=2×10−4\alpha^2 = \frac{2 \times 10^{-5}}{0.1} = 2 \times 10^{-4}
    α=2×10−4=2×10−2\alpha = \sqrt{2 \times 10^{-4}} = \sqrt{2} \times 10^{-2}
    α≈1.414×10−2\alpha \approx 1.414 \times 10^{-2}

    Let's check the approximation. If α=1.414×10−2=0.01414\alpha = 1.414 \times 10^{-2} = 0.01414, then 1−α=0.985861-\alpha = 0.98586, which is close to 1. So the approximation is valid.

  6. Express α\alpha in the required format:
    α=1.414×10−2=14.14×10−3\alpha = 1.414 \times 10^{-2} = 14.14 \times 10^{-3}
    The problem asks for α=a×10−3\alpha = a \times 10^{-3}.
    So, a=14.14a = 14.14.

  7. Nearest integer value of a:
    The nearest integer to 14.14 is 14.

The final answer is 14\boxed{14}.