Question

from the following molar conductivities at infinite dilution, ^m at infinte for Al2(SO4)3 = 858 cm2mol-1. ^m at infinite for NH4OH = 238.3 S cm2 mol-1 ^m at infinity for (NH4)2SO4= 238.4 S cm2 mol-1.. Calculate ^m infinite for Al(OH)3

Answer 1

Answer:

786.3 Scm

2

mol

−1

Explanation:

Molar conductivities at infinite dilution :

Λ

m

0

for Al

2

(SO

4

)

3

=858 Scm

2

mol

−1

Λ

m

0

for NH

4

OH=238.3 Scm

2

mol

−1

Λ

m

0

for (NH

4

)

2

SO

4

=238.4 Scm

2

mol

−1

Al

2

(SO

4

)

3

⇌2Al

3+

+3SO

4

2−

NH

4

OH⇌NH

4

+

+OH

−

(NH

4

)

2

SO

4

⇌2NH

4

+

+SO

4

2−

Al(OH)

3

⇌Al

3+

+3OH

−

2(Λ

m

0

Al(OH)

3

)=Λ

m

0

Al

2

(SO

4

)

3

+6(Λ

m

0

NH

4

OH)−3(Λ

m

0

(NH

4

)

2

SO

4

)

2(Λ

m

0

Al(OH)

3

)=858+6(238.3)−3(238.4)

=1572.6

Λ

m

0

Al(OH)

3

=

2

1572.6

=786.3 Scm

2

mol

−1