What is the concentration of hydrogen ions
in a solution containing 0.645 M HOCN
(cyanic acid) and 0.235 M (NaOCN) (sodium
cyanate)? The ionization constant of cyanic
acid is 3.5 × 10−4
.
Answer in units of M
To determine the concentration of hydrogen ions in the given solution, we first need to set up the dissociation reaction for cyanic acid (HOCN):
HOCN ⟶ H+ + OCN-
The ionization constant (Ka) for cyanic acid is given as 3.5 × 10^-4. This can be expressed as:
Ka = [H+][OCN-] / [HOCN]
Since the concentration of sodium cyanate (NaOCN) is not considered to affect the hydrogen ion concentration, we only need to focus on the concentration of HOCN. Thus, the equation becomes:
Ka = [H+][OCN-] / [HOCN]
3.5 × 10^-4 = [H+] * [OCN-] / 0.645
To find the concentration of hydrogen ions ([H+]), we rearrange the equation:
[H+] = (Ka * [HOCN]) / [OCN-]
[H+] = (3.5 × 10^-4) * (0.645) / 0.235
Calculating this gives us:
[H+] ≈ 9.59 × 10^-4 M
Therefore, the concentration of hydrogen ions in the given solution is approximately 9.59 × 10^-4 M.
To determine the concentration of hydrogen ions in the solution, we need to calculate the concentration of cyanic acid that dissociates into hydrogen ions.
The ionization reaction of cyanic acid (HOCN) is as follows:
HOCN <--> H+ + OCN-
The ionization constant, also known as the acid dissociation constant (Ka), is given as 3.5 × 10^(-4). This is an equilibrium constant, expressed as the ratio of the concentrations of the products (H+ and OCN-) to the concentration of the reactant (HOCN).
Using the given concentrations of HOCN and NaOCN, we can assume that the NaOCN does not contribute any hydrogen ions to the solution since it is a salt and does not dissociate completely.
Let's denote the concentration of HOCN that ionizes as 'x'. After ionization, the concentration of H+ and OCN- would also be 'x'. Therefore, at equilibrium, the concentration of HOCN remaining will be (0.645 M - x) and the concentration of the products H+ and OCN- will be 'x'.
Now we can set up an expression for the ionization constant:
Ka = [H+][OCN-] / [HOCN]
Substituting the known values:
3.5 × 10^(-4) = x * x / (0.645 - x)
Since Ka is a small value, we can assume that 'x' is significantly less than 0.645, so 'x' can be ignored in the denominator.
3.5 × 10^(-4) = x^2 / 0.645
Cross-multiplying:
x^2 = 0.645 * 3.5 × 10^(-4)
x^2 = 2.2575 × 10^(-4)
Taking the square root of both sides:
x = √(2.2575 × 10^(-4))
x ≈ 0.01593
So, the concentration of hydrogen ions (H+) in the solution is approximately 0.01593 M.