HCN = CN- + H+
DrBob222
answered
I.........4.68E-3....0.....0
C..........-x........x......x
E........4.69E-4-x...x......x
Ka = (H^+)(CN^-)/(HCN)
Ka = (x)(x)/(4.68E-4-x)
You know x = 1.51E-6. Substitute into the above and solve for Ka.
Explain Bot
answered
The equation is given as:
pH = pKa + log([A-]/[HA])
Where:
- pH is the negative logarithm of the hydrogen ion concentration ([H+])
- pKa is the negative logarithm of the acid dissociation constant (Ka) for HCN
- [A-] is the concentration of the conjugate base CN-
- [HA] is the concentration of the acid HCN
In this problem, we are given the concentration of [H+] as 0.00000151 M, which is equal to [HA] since HCN is a weak acid. We need to find the value of pKa to estimate the equilibrium constant.
To find the pKa, we rearrange the Henderson-Hasselbalch equation as:
pKa = pH - log([A-]/[HA])
The concentration of CN- ([A-]) can be determined by subtracting the concentration of [H+] from the initial concentration of HCN:
[A-] = initial concentration of HCN - [H+] = 4.68×10-3 M - 0.00000151 M
Now we have all the necessary values to calculate the pKa:
pKa = -log(0.00000151 / (4.68×10-3 - 0.00000151))
Next, we can use the relationship between pKa and Ka:
Ka = 10^(-pKa)
Finally, to estimate the equilibrium constant (K) for the dissociation of HCN, we can use the expression:
K = [CN-][H+] / [HCN]
Since the concentration of CN- ([A-]) is equal to [HCN], we can simplify the equation to:
K = [H+]^2 / [HCN]
Substituting the known values, we can now estimate the equilibrium constant K.
