The pH of 0.15 mol/L of hydrocyanic acid is 5.07. Calculate Ka of hydrocyanic acid.
Convert pH to (H^+). It's approximately, but you need to do it exactly, 9E-6
...............HCN ==> H^+ + CN^-
initial........0.15M...0.......0
change........-1E-6...1E-6...1E-6
equil.......0.15-1E-6..1E-6...1E-6
Ka = (H^+)(CN^-)/(HCN)
Substitute from the ICE chart and solve for Ka.
How do you calculate the [H^+] and where did you get the -1E-6 from?
pH = -log(H^+)
5.07 = -log(H^+)
-5.07 = log(H^+)
(H^+) = 8.51E-6
Note that at the top of the page I estimated the (H^+) to be 9E-6M and noted that you should calculate it more accurately. Then I substituted into the ICE chart and I used 1E-6. That was a typo but since it was just an estimate it doesn't make much difference. You should substitute 8.51E-6 for (H^+).
To calculate the Ka (acid dissociation constant) of hydrocyanic acid (HCN), we need to first understand the acid-base equilibrium reaction it undergoes in water:
HCN (aq) + H2O (l) ⇌ H3O+ (aq) + CN- (aq)
The equilibrium expression for this reaction is given by:
Ka = [H3O+][CN-] / [HCN]
However, we are only given the pH of the hydrocyanic acid solution. The pH is a measure of the concentration of hydronium ions (H3O+) in a solution. Thus, we need to determine the concentration of hydronium ions (H3O+) in order to calculate the Ka value.
In this case, the pH of the hydrocyanic acid solution is given as 5.07. To determine the concentration of hydronium ions (H3O+), we can use the pH equation:
pH = -log[H3O+]
Re-arranging the equation, we get:
[H3O+] = 10^(-pH)
Now, substituting the given pH value into the equation, we find:
[H3O+] = 10^(-5.07)
[H3O+] = 1.07 x 10^(-6) mol/L
Since hydrocyanic acid (HCN) is a weak acid, it does not fully ionize in water. Therefore, we can assume that the moles of hydrocyanic acid (HCN) that dissociate to form hydronium ions (H3O+) and cyanide ions (CN-) are equal.
Let's assume the concentration of dissociated HCN is "x" (in mol/L). Therefore, the concentration of H3O+ and CN- ions is also "x" (in mol/L).
Using the given initial concentration of hydrocyanic acid (0.15 mol/L), we can express the equilibrium concentrations in terms of "x" as follows:
[H3O+] = x
[CN-] = x
[HCN] = 0.15 - x
Substituting these concentrations into the equilibrium expression for Ka, we get:
Ka = [H3O+][CN-] / [HCN]
Ka = (x)(x) / (0.15 - x)
Since we know that [H3O+] = 1.07 x 10^(-6) mol/L, we can substitute this value into the expression for Ka:
Ka = (1.07 x 10^(-6))(1.07 x 10^(-6)) / (0.15 - 1.07 x 10^(-6))
Now, we can calculate the value of Ka using a calculator or a software tool. The calculated value will be the Ka of hydrocyanic acid.