What is the pH of a solution that is 0.0100 M in HCN and 0.0200 M in NaCN?
What is the pH of a solution that is 0.0100 M in HCN and 0.0200 M in NaCN?
To determine the pH of a solution that contains a weak acid and its conjugate base, you need to consider the dissociation of the acid and the equilibrium of the reaction.
In this case, HCN is a weak acid, while NaCN is the corresponding conjugate base (sodium cyanide). The dissociation of HCN in water can be represented by the following equation:
HCN (aq) ⇌ H+ (aq) + CN- (aq)
The equilibrium constant for this dissociation is known as the acid dissociation constant (Ka) and can be expressed as:
Ka = [H+][CN-] / [HCN]
Since HCN and CN- are present in the solution, they can react with water and contribute to the concentration of H+ ions in the solution. However, because CN- is a strong enough base, it will react with water to a lesser extent than water itself, resulting in the formation of OH- ions.
To determine the pH, you need to calculate the concentration of H+ ions in the solution. One way to do this is to use the equation:
[H+] = √(Ka x [HCN] / [CN-])
First, calculate the value of Ka for HCN. The Ka value for HCN is 6.2 x 10^-10 (you can look this up in a table or a chemistry reference).
Next, plug in the values:
[H+] = √(6.2 x 10^-10 x 0.0100 M / 0.0200 M)
Simplifying the equation gives:
[H+] = √(3.1 x 10^-12)
[H+] = 1.76 x 10^-6 M
Finally, calculate the pH by taking the negative logarithm of the concentration of H+ ions:
pH = -log10(1.76 x 10^-6)
pH ≈ 5.75
Therefore, the pH of the solution is approximately 5.75.