What is the PH of a 0.747M sodium cyanide solution (NaCN) knowing that the Ka of Hydrocyanic acid is 4.9x10-10?
The pH is determined by the hydrolysis of the CN^- as follows:
.....................CN^- + HOH ==>HCN + OH^-
I..................0.747.......................0..........0
C...................-x...........................x...........x
E.................0.747-x.....................x...........x
Kb (for CN^-) = Kw/Ka(for HCN) = (HCN)(OH^-)/(CN^-)
Plug the E line into the Kb expression and solve for x = (OH^-). Convert to (H^+) and pH. Post your work if you get stuck.
To find the pH of a sodium cyanide (NaCN) solution, we need to consider the hydrolysis reaction of the cyanide ion (CN-) with water. The hydrolysis reaction can be represented as follows:
CN- + H2O ⇌ HCN + OH-
Given that the Ka of hydrocyanic acid (HCN) is 4.9x10^-10, we can use the Ka expression to determine the concentration of HCN formed:
Ka = [HCN][OH-] / [CN-]
Since the concentration of OH- can be assumed to be 0.0 for a dilute solution of NaCN, we can simplify the expression to:
Ka = [HCN] / [CN-]
Now, let's determine the concentration of HCN. The concentration of HCN can be assumed to be the same as the concentration of CN- (0.747 M) because NaCN dissociates completely in water to form Na+ and CN-. Thus, [HCN] = 0.747 M.
Plugging the values into the expression, we get:
4.9x10^-10 = 0.747 / [CN-]
Rearranging the equation to solve for [CN-], we have:
[CN-] = 0.747 / 4.9x10^-10
[CN-] ≈ 1.524x10^9 M
Since the hydrolysis reaction produces equal amounts of HCN and OH-, and the OH- concentration can be neglected, the concentration of HCN is also approximately 1.524x10^9 M.
To calculate the pH, we need to consider the dissociation of HCN (a weak acid) in water:
HCN + H2O ⇌ H3O+ + CN-
The equilibrium expression for this reaction is given by:
Ka = [H3O+][CN-] / [HCN]
Since the concentration of CN- (1.524x10^9 M) and HCN (1.524x10^9 M) are equal, we can simplify the expression to:
Ka = [H3O+] / [HCN]
Let's rearrange the equation to solve for [H3O+]:
[H3O+] = Ka × [HCN]
[H3O+] = (4.9x10^-10) × (1.524x10^9 M)
[H3O+] ≈ 7.4636
The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration, [H3O+]. Therefore:
pH ≈ -log10(7.4636)
pH ≈ 0.1275
Hence, the pH of a 0.747 M sodium cyanide (NaCN) solution is approximately 0.1275.