What is the concentration of hydrogen ions in a solution containing 0.785 M HOCN (cyanic acid) and 0.659 M (NaOCN) (sodium cyanate)? The ionization constant of cyanic acid is 3.5 × 10−4.
Answer in units of M
The HH equation again.
To determine the concentration of hydrogen ions in the solution, we need to consider the ionization of cyanic acid (HOCN) and the dissociation of sodium cyanate (NaOCN):
HOCN ⇌ H+ + OCN-
NaOCN ⇌ Na+ + OCN-
Since we are given the concentrations of HOCN and NaOCN, we can assume that the cyanic acid and sodium cyanate have dissociated completely into their respective ions.
First, we need to calculate the concentration of H+ ions from HOCN.
The ionization constant of cyanic acid (HOCN) is given as 3.5 × 10^-4. This constant is the ratio of the concentrations of the products (H+ and OCN-) to the concentration of the reactant (HOCN).
So we have:
[H+][OCN-] / [HOCN] = 3.5 × 10^-4
We are given the concentration of HOCN as 0.785 M. Let's represent the concentration of H+ ions as x, then the concentration of OCN- ions will also be x (as both are produced in a 1:1 ratio).
Substituting these values into the equation, we get:
(x)(x) / 0.785 = 3.5 × 10^-4
Simplifying the equation:
x^2 = (3.5 × 10^-4)(0.785)
x^2 = 2.7475 × 10^-4
x = sqrt(2.7475 × 10^-4)
Now, let's calculate the concentration of H+ ions from NaOCN.
Since sodium cyanate (NaOCN) dissociates into Na+ and OCN- ions, the concentration of H+ ions will be 0 M (as there is no direct production of H+ ions).
Therefore, the concentration of H+ ions in the solution is solely determined by the ionization of HOCN.
Plugging in the values into a calculator, we find:
x ≈ 0.01657 M
Therefore, the concentration of hydrogen ions in the solution is approximately 0.01657 M.