Home
Class 12
CHEMISTRY
Calculate percentage hydrolysis in 0.00...

Calculate percentage hydrolysis in `0.003` M aqueous solution `K_(a)` for HOCN = `3.35 xx 10^(-4) `and `K_(w) = 10^(-14)`

Text Solution

AI Generated Solution

To calculate the percentage hydrolysis of HOCN in a 0.003 M aqueous solution, we will follow these steps: ### Step 1: Write the hydrolysis reaction The hydrolysis of HOCN (cyanic acid) can be represented as: \[ \text{HOCN} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OCN}^- \] ### Step 2: Identify the given values - Concentration of HOCN, \( C = 0.003 \, \text{M} \) ...
Promotional Banner

Similar Questions

Explore conceptually related problems

a. Calculate the percentage hydrolysis of 0.003M aqueous solution of NaOH. K_(a) for HOCN = 3.3 xx 10^(-4) . b. What is the pH and [overset(Theta)OH] of 0.02M aqueous solution of sodium butyrate. (K_(a) = 2.0 xx 10^(-5)) .

The degree of hydrolysis of 0.005 M aqueous solution of NaOCN will be ( K_a for HOCN = 4.0xx10^-4 )

What is the percentage hydrolysis of CH_3COONa in 0.01 M Solution ? (K_a= 1.78 xx 10 ^(-5) and K_w = 1.0 xx 10 ^(-14))

The pH of 0.5 M aqueous solution of HF (K_(a)=2xx10^(-4)) is

Calculate the percentage hydrolysis of decinormal solution of ammonium acetate given that k_(a) = 1.75 xx 10^(-5), K_(b) =1.80 xx 10^(-5) and K_(w) = 1.0 xx 10^(-14)

Calculate the percentage hydrolysis of decinormal solution of ammonium acetate given that k_(a) = 1.75 xx 10^(-5), K_(b) =1.80 xx 10^(-5) and K_(w) = 1.0 xx 10^(-14)

Calculate the degree of hydrolysis of 0.1 M solution of sodium acetate at 298 K : K_(a) = 1.8 xx 10^(-5) .

Calculate the degree of hydrolysis and pH of 0.2M solution of NH_(4)C1 Given K_(b) for NH_(4)OH is 1.8 xx 10^(-5) .

Calculate the solubility of CaF_(2) in a solution buffered at pH = 3.0. K_(a) for HF = 6.3 xx 10^(-4) and K_(sp) of CaF_(2) = 3.45 xx 10^(-11) .

Calculate the [OH^(-)] in 0.01M aqueous solution of NaOCN(K_(b) for OCN^(-)=10^(-10)) : (a) 10^(-6) M (b) 10^(-7) M (c) 10^(-8) M (d)None of these