JEE MAIN - Chemistry (2007 - No. 28)
The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move
freely through a solution) at 25oC are given below:
$$ \wedge _{C{H_3}COONa}^o$$ = 91.0 S cm2/equiv
$$ \wedge _{HCl}^o$$ = 426.2 S cm2/equiv
What additional information/quantity one needs to calculate $$ \wedge ^o$$ of an aqueous solution of acetic acid?
$$ \wedge _{C{H_3}COONa}^o$$ = 91.0 S cm2/equiv
$$ \wedge _{HCl}^o$$ = 426.2 S cm2/equiv
What additional information/quantity one needs to calculate $$ \wedge ^o$$ of an aqueous solution of acetic acid?
$$ \wedge ^o$$ of chloroacetic acid (C/CH2COOH)
$$ \wedge ^o$$ of NaCl
$$ \wedge ^o$$ of CH3COOK
The limiting equivalent conductance of $${H^ + }( \wedge _{{H^ + }}^o)$$
Explanation
NOTE : According to Kohlrausch's law, molar conductivity of weak electrolyte acetic acid $$\left( {C{H_3}COOH} \right)$$ can be calculated as follows:
$${\Lambda ^o}_{C{H_3}COOH} = \left( {{\Lambda ^o}_{C{H_3}COONa} + {\Lambda ^o}_{HCl}} \right) - {\Lambda ^o}_{NaCl}$$
$$\therefore$$ Value of $${\Lambda ^o}_{NaCl}$$ should also be known
for calculating value of $${\Lambda ^o}_{C{H_3}COOH}$$
$${\Lambda ^o}_{C{H_3}COOH} = \left( {{\Lambda ^o}_{C{H_3}COONa} + {\Lambda ^o}_{HCl}} \right) - {\Lambda ^o}_{NaCl}$$
$$\therefore$$ Value of $${\Lambda ^o}_{NaCl}$$ should also be known
for calculating value of $${\Lambda ^o}_{C{H_3}COOH}$$
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