What is the pH of a 0.0250 M solution of HCN? Ka = 4.0x10^-10
HCN >< H+ CN-
.250-x <> x+ x
4E-10=x^2/(.250-x) appx x^2/.250
[H+]= sqrt (4*.250E-10)
= 1E-5
pH= 5
To determine the pH of a solution of HCN, we need to calculate the concentration of H+ ions in the solution.
HCN is a weak acid, so it will undergo a partial dissociation in water. The dissociation reaction is as follows:
HCN + H2O ⇌ H3O+ + CN-
The equilibrium constant for this reaction is given by the acid dissociation constant (Ka), which is 4.0x10^-10.
We can set up an ICE (Initial, Change, Equilibrium) table to help us solve for the concentration of H+ ions at equilibrium.
Let x be the concentration of H+ ions at equilibrium. Since HCN dissociates in a 1:1 ratio, the concentration of CN- ions at equilibrium will also be x.
HCN + H2O ⇌ H3O+ + CN-
Initial: 0.0250 0 0
Change: -x -x +x
Equilibrium: (0.0250 - x) (-x) (x)
Using the equilibrium concentrations, we can write the expression for Ka:
Ka = [H3O+][CN-] / [HCN]
Substituting the equilibrium concentrations, we have:
4.0x10^-10 = x * x / (0.0250 - x)
Since x is small compared to 0.0250, we can approximate (0.0250 - x) as 0.0250:
4.0x10^-10 = x^2 / 0.0250
Rearranging the equation:
x^2 = (4.0x10^-10) * 0.0250
Solving for x:
x^2 = 1.0x10^-11
x = √(1.0x10^-11)
x ≈ 1.0x10^-6
The concentration of H+ ions (and therefore also the concentration of H3O+ ions) is approximately 1.0x10^-6 M.
Finally, we can calculate the pH using the formula:
pH = -log[H3O+]
pH = -log(1.0x10^-6)
pH ≈ 6.00
Therefore, the pH of the 0.0250 M solution of HCN is approximately 6.00.