What is the concentration oh hydrogen ions in a solution containing .676 M HOCN and .413 M NaOCN? The ionization constant of cyanic acid is 3.5 x 10^-4. Answer in units of M.
To find the concentration of hydrogen ions in the solution, we need to first determine the concentration of cyanate ions (OCN-) in the solution.
The cyanic acid (HOCN) dissociates in water according to the following equilibrium reaction:
HOCN ⇌ H+ + OCN-
The ionization constant (Ka) is given as 3.5 x 10^-4, which is the ratio of the products (H+ and OCN-) to the reactant (HOCN) concentrations. Mathematically, it is expressed as:
Ka = [H+][OCN-] / [HOCN]
Since the concentration of HOCN is given as 0.676 M and the concentration of NaOCN is given as 0.413 M, we can assume that all NaOCN has dissociated into OCN-. Therefore, the concentration of OCN- is 0.413 M.
Now, let's rearrange the equilibrium expression to solve for the concentration of hydrogen ions ([H+]):
[H+] = (Ka * [HOCN]) / [OCN-]
Substituting the known values:
[H+] = (3.5 x 10^-4 * 0.676 M) / 0.413 M
Simplifying the expression:
[H+] = 5.71 x 10^-4 M
Therefore, the concentration of hydrogen ions in the solution is 5.71 x 10^-4 M.