Calculate the percent hydrolysis in a 0.0100 M solution of KCN .( K a =6.2 × 10-10) write answer up to two digits after decimal place.

Question

Calculate the percent hydrolysis in a 0.0100 M solution of KCN .( K a =6.2 × 10-10) write answer up to two digits after decimal place. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

K h =(10-14/6.2 × 10-10)=(1/6.2) × 10-4 ( K h / c ) =1.6 × 10-3 Assuming h < < 1 h 2=( K h / c ) h =0.04 Percent hydrolysis =4 %

Q. Calculate the percent hydrolysis in a $0.0100 M$ solution of $K CN . (K_{a} = 6.2 \times 1 0^{- 10})$ write answer up to two digits after decimal place.

$K_{h} = \frac{1 0 ^{- 14}}{6.2 \times 1 0 ^{- 10}} = \frac{1}{6.2} \times 1 0^{- 4}$
$\frac{K _{h}}{c} = 1.6 \times 1 0^{- 3}$
Assuming $h << 1$
$h^{2} = \frac{K _{h}}{c}$
$h = 0.04$
Percent hydrolysis $= 4%$

Q. Calculate the percent hydrolysis in a 0.0100M solution of KCN.(Ka​=6.2×10−10) write answer up to two digits after decimal place.

Kh​=6.2×10−1010−14​=6.21​×10−4 cKh​​=1.6×10−3 Assuming h<<1 h2=cKh​​ h=0.04 Percent hydrolysis =4%

Q. Calculate the percent hydrolysis in a $0.0100 M$ solution of $K CN . (K_{a} = 6.2 \times 1 0^{- 10})$ write answer up to two digits after decimal place.

$K_{h} = \frac{1 0 ^{- 14}}{6.2 \times 1 0 ^{- 10}} = \frac{1}{6.2} \times 1 0^{- 4}$
$\frac{K _{h}}{c} = 1.6 \times 1 0^{- 3}$
Assuming $h << 1$
$h^{2} = \frac{K _{h}}{c}$
$h = 0.04$
Percent hydrolysis $= 4%$