What is percentage of hydrolysis in 0.003M aqueous solution of  NaOCN  (Ka for  HOCN=3.33X10-4)

To find the percentage of hydrolysis for a 0.003M aqueous solution of NaOCN, we first need to understand that NaOCN is a salt of a weak acid (HOCN) and a strong base (NaOH). When NaOCN is dissolved in water, it will undergo hydrolysis to form its parent acid (HOCN) and base (OH-).

The hydrolysis reaction is as follows:

OCN- + H2O ⇌ HOCN + OH-

The degree of hydrolysis (h) can be calculated using the formula:

h = √(Ka/Kw) / C

where: Ka is the acid dissociation constant of the weak acid (HOCN), which is given as 3.33x10^-4. Kw is the ion product of water, which at 25°C is 1.0x10^-14. C is the concentration of the salt (NaOCN), which is given as 0.003M.

Substituting the given values into the formula, we get:

h = √((3.33x10^-4) / (1.0x10^-14)) / 0.003 h = √(3.33x10^10) / 0.003 h = √(3.33) / 0.003 h = 0.0576

To find the percentage of hydrolysis, we multiply h by 100:

Percentage of hydrolysis = h * 100 Percentage of hydrolysis = 0.0576 * 100 Percentage of hydrolysis = 5.76%

So, the percentage of hydrolysis in a 0.003M aqueous solution of NaOCN is approximately 5.76%.