What is the percentage hydrolysis of NaCN in N/80 solution,when the dissociation constant for HCN is 1.3 × 10-9 and Kw = 1.0 × 10-14

Question

What is the percentage hydrolysis of NaCN in N/80 solution,when the dissociation constant for HCN is 1.3 × 10-9 and Kw = 1.0 × 10-14 (A) 2.48 (B) 5.26 (C) 8.2 (D) 9.6

Tardigrade Admin · Accepted Answer

h=√(Kw/KaC)=√(10-14/1.3×10-9 (1 /80)) =√(80×10-5/1.3)=√(80/13)×10-4 =2.48×10-2 =2.48 %

Q. What is the percentage hydrolysis of NaCN in N/80 solution,when the dissociation constant for HCN is $1.3 \times 1 0^{- 9}$ and $K_{w} = 1.0 \times 1 0^{- 14}$

2.48

5.26

8.2

9.6

$h = \frac{K _{w}}{K _{a} C} = \frac{1 0 ^{- 14}}{1.3 \times 1 0 ^{- 9} ( 1/80 )}$
$= \frac{80 \times 1 0 ^{- 5}}{1.3} = \frac{80}{13} \times 1 0^{- 4}$
$= 2.48 \times 1 0^{- 2}$
$= 2.48%$

Q. What is the percentage hydrolysis of NaCN in N/80 solution,when the dissociation constant for HCN is 1.3×10−9 and Kw​=1.0×10−14

2.48

5.26

8.2

9.6

h=Ka​CKw​​​=1.3×10−9(1/80)10−14​​ =1.380×10−5​​=1380​×10−4​ =2.48×10−2 =2.48%

Q. What is the percentage hydrolysis of NaCN in N/80 solution,when the dissociation constant for HCN is $1.3 \times 1 0^{- 9}$ and $K_{w} = 1.0 \times 1 0^{- 14}$

$h = \frac{K _{w}}{K _{a} C} = \frac{1 0 ^{- 14}}{1.3 \times 1 0 ^{- 9} ( 1/80 )}$
$= \frac{80 \times 1 0 ^{- 5}}{1.3} = \frac{80}{13} \times 1 0^{- 4}$
$= 2.48 \times 1 0^{- 2}$
$= 2.48%$