Calculate the pH of a 5.75×10-2 M solution of the weak acid HCN (make an approximate calculation assuming that initial concentration is equal to the equilibrium concentration). Round your answer to 3 significant digits.
HCN = CN- + H+
See your HClO question.
To calculate the pH of a weak acid solution like HCN, we need to know the concentration of the acid and the acid dissociation constant, also known as the Ka.
The equilibrium expression for the dissociation of HCN is:
HCN ⇌ CN- + H+
The Ka value for HCN can be found in a reference table, which is 4.9 × 10^-10.
Now, we need to set up an ICE table (Initial, Change, Equilibrium) to calculate the concentrations at equilibrium. Since the initial concentration is equal to the equilibrium concentration in this case, we only need to consider the change.
Let x represent the concentration of H+ and CN- at equilibrium.
HCN ⇌ CN- + H+
Initial: 5.75×10^-2 0 0
Change: -x +x +x
Equilibrium: 5.75×10^-2 - x x x
Since HCN is a weak acid, we can assume that x is small compared to the initial concentration. Thus, we can approximate the equilibrium concentration of HCN as 5.75×10^-2.
With this approximation, the equilibrium concentration of CN- and H+ will also be approximately 5.75×10^-2.
Now, we can write the expression for Ka and solve for x:
Ka = [CN-][H+]/[HCN] = (x)(x)/(5.75×10^-2 - x)
Since x is small, we can assume that 5.75×10^-2 - x is approximately equal to 5.75×10^-2.
Plugging in the values, the equation becomes:
4.9 × 10^-10 = (x)(x)/(5.75×10^-2)
Rearranging the equation and solving for x, we get:
x^2 = (4.9 × 10^-10)(5.75×10^-2)
x^2 = 2.8175 x 10^-12
x ≈ 1.677 x 10^-6
Now that we have the concentration of H+, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(1.677 x 10^-6)
pH ≈ 5.774
Rounding to 3 significant digits, the pH of the 5.75×10^-2 M solution of HCN is approximately 5.77.