Question

Calculate molar conductance for $N{H_4}OH$ , given that molar conductance for $Ba{(OH)_2}$ , $BaC{l_2}$ , $N{H_4}Cl$ are $523.28\,,\,\,280.0\,\,and\,\,129.8\,S\,c{m^2}\,e{q^{ - 1}}$ respectively.

Answer 1

For calculating the molar conductance for weak electrolytes, we use the Kohlraush’s law try to make formula for ammonium hydroxide using all other chemical formula given like barium hydroxide, barium chloride and ammonium chloride. You have to do the same changes with their molar conductance value.

Complete step-by-step answer: Molar conductance is the conductance of one mole of a solution when it is dissolved in water. During dissolution of the particles they start conducting electricity and give a particular value on the potentiometer which is used for measuring potential difference. As we know and as we read in our junior classes that there are two types of electrolytes: strong and weak electrolytes. In case of strong electrolyte the solution gets dissociated in full extent while in weak electrolytes dissociation is low. Thus when we want to calculate the molar conductance, we used to plot a graph for strong electrolytes. But for weak electrolytes we use Kohlrausch law, by using molar conductance of the strong electrolyte we can find out the molar conductance of weak one.
$Ba{(OH)_2} = \,B{a^{ + 2}}\, + \,2\,O{H^ - }\,\left( {{\Lambda _m} = \,523.28} \right)$
$BaC{l_2} = \,B{a^{ + 2}}\, + \,2\,C{l^ - }\,\,\,\,\left( {{\Lambda _m} = \,280.0} \right)$
$N{H_4}Cl\, = \,NH_4^ + \, + \,C{l^ - }\,\,\,({\Lambda _m} = \,129.8)$
Here, we have three strong electrolytes whose molar conductance is given. Try to make a formula of ammonium hydroxide using the formula of others.
$\left[ {2\,[\,NH_4^ + \, + \,C{l^ - }]\, + \,[B{a^{ + 2}}\, + \,2O{H^ - }]} \right]\, - [B{a^{ + 2}}\, + \,2C{l^ - }]$
$2(N{H_4}OH)\, + \,BaC{l_2}\, - \,BaC{l_2}$
By adding ammonium chloride and barium hydroxide, we have made the formula of ammonium hydroxide but there is barium chloride leave at the end, so to make it disappear or to finish it, we have to subtract the barium chloride formula from the additive part. At last we will get the ammonium hydroxide. We can write it in the form of Kohlrausch law as, $\,\left[ {2\Lambda _{N{H_4}Cl}^ \circ \, + \,\Lambda _{Ba{{(OH)}_2}}^ \circ } \right] - \,\left[ {\Lambda _{BaC{l_2}}^ \circ } \right] = 2\Lambda _{N{H_4}OH}^ \circ$
Now putting the value of molar conductance for strong electrolytes,
$\,\left[ {2 \times 129.8\, + \,523.28} \right] - \,\left[ {280.0} \right] = 2\Lambda _{N{H_4}OH}^ \circ$
$2\Lambda _{N{H_4}OH}^ \circ = \,502.88$
$\Lambda _{N{H_4}OH}^ \circ = 251.44\,S\,c{m^2}\,e{q^{ - 1}}$
Thus molar conductance of ammonium hydroxide is $251.44\,S\,c{m^2}\,e{q^{ - 1}}$ .

Note: The Kohlraush’s law is in which we use molar conductance of strong electrolyte for determining the molar conductance of weak electrolytes. As above we firstly make formula of weak electrolyte, make use of ions formed after dissociation of these strong electrolyte, then add them properly to form weak electrolyte. There are many cases where you will get the formula simply by adding the strong electrolyte but as above we have to multiply the molar conductance of ammonium chloride to maintain the stereochemistry.