The PH of an aqueous solution of NaCN is 10.38. What is [CN-] in this solution?
Sorry for double posting but I got a solution of [CN-] = 0.0036 M. Can someone verify if that is right.
Can you show the working. Please
Using 6.2 x 10^-10 for Ka for HCN, I obtained 0.00357 M for (CN^-) which rounds to 0.0036. I think you probably worked it correctly.
To find the concentration of CN- in the solution, we need to revert the pH value back to the concentration of H+ ions.
Step 1: Determine the concentration of H+ ions (H3O+) using the pH value.
In the case of a basic solution (pH > 7), we use the equation:
pOH + pH = 14
Since pOH = 14 - pH, we can calculate the pOH value as 14 - 10.38 = 3.62.
Step 2: Calculate the OH- concentration using the pOH value.
To get the concentration of OH-, we use the equation:
pOH = -log[OH-]
Rearranging the equation, we have:
[OH-] = 10^(-pOH)
Plugging in the pOH value, we find [OH-] = 10^(-3.62).
Step 3: Calculate the concentration of CN-.
Since NaCN is a salt, it dissociates completely in water. This means that for every NaCN molecule that dissolves, it produces one Na+ ion and one CN- ion. Therefore, [CN-] is equal to [OH-], so [CN-] = 10^(-3.62).
Using scientific notation, the concentration of CN- in the solution is approximately [CN-] = 3.85 x 10^(-4) M.