The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If λ X - 0 ≈ λ Y - 0 , the difference in their p K a values, p K a H X - p K a ( H Y ) , is (Consider degree of ionization of…

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The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If &#955; X - 0 &#8776; &#955; Y - 0 , the difference in their p K a values, p K a H X - p K a ( H Y ) , is (Consider degree of ionization of both acids to << 1)

Aicharya · Accepted Answer

&#955; X - 0 &#8776; &#955; Y - 0 &#8658; &#955; H + 0 + &#955; X - 0 &#160; &#8776; &#955; H + 0 + &#955; Y - 0 &#8658; &#955; H X 0 &#8776; &#955; H Y 0 ...... (i) Also, &#955; m &#955; m 0 = &#945; , So &#955; m H X = &#160; &#955; m 0 &#945; 1 and &#955; m H Y = &#160; &#955; m 0 &#945; 2 (Where &#945; 1 and &#945; 2 are degree of dissociation of HX and HY respectively) Now, given that &#955; m H Y = 10 &#955; m H X &#8658; &#955; m 0 &#945; 2 = 10 &#215; &#955; m 0 &#945; 1 &#945; 2 = 10 &#945; 1 ...... (ii) K a = C &#945; 2 1 - &#945; , But &#945; &#8810; 1 , therefore K a = C &#945; 2 &#8658; K a ( H X ) K a ( H Y ) = 0.01 &#945; 1 2 0.1 &#945; 2 2 = 0.01 0.1 &#215; 1 10 2 = 1 1000 &#8658; [ log &#8289; ( K a H X - log &#8289; K a H Y = - 3 &#8658; p K a H X - p K a H Y = 3

The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar…

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