Final answer:
The pH of a 1.23 M solution of NaCN is approximately 0.35.
Step-by-step explanation:
The pH of a 1.23 M solution of NaCN can be calculated using the equation: pH = -log[H+].
First, we need to calculate the concentration of H+ ions in the solution. NaCN is a salt that dissociates in water to produce Na+ and CN- ions. CN- can act as a weak base and combine with water to form HCN and OH- ions. Since NaCN is a strong electrolyte, we can assume complete dissociation, and the concentration of CN- ions will be the same as NaCN.
To find the concentration of H+ ions, we need to use the equilibrium constant (Kw) of water to calculate the concentration of OH-. Kw = [H+][OH-] = 1.0 x 10^-14. Assuming complete dissociation of NaCN, the concentration of CN- ions will be 1.23 M. Since every CN- ion combines with a water molecule to form one OH- ion, the concentration of OH- ions will also be 1.23 M.
Now, we can calculate the concentration of H+ ions using the equation [H+][OH-] = 1 x 10^-14. [H+][1.23] = 1 x 10^-14. [H+] = 1 x 10^-14 / 1.23 = 8.13 x 10^-15 M.
Finally, we can calculate the pH using the equation pH = -log[H+]. pH = -log(8.13 x 10^-15) = 14 - (-15log(8.13)). pH ≈ 14 + 15(-0.91) = 14 - 13.65 = 0.35.