What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10^-10?
I am not sure how to this question... I did 6.2x10^-10/ 1.24?? I am not sure if it is right OR are we suppose to find the Kb value which i already did Kb= kw/ka = 1.0x10^-14/6.2x10^-10 = 1.61x10^-5 then divide it by 1.24 = 1.30X10^-5???
HCN is an acid. Treat it as an acid.
HCN==> H^+ + CN^-
Ka = (H^+)(CN^-)/(HCN) = 6.2 x 10^-10
Now do the ICE table.
I = initial concns:
C = change in concns:
E = equilibrium concns:
Initial (before any ionization takes place):
(HCN) = 1.24
(H^+) = 0
(CN^-) = 0
change:
(H^+) = +y
(CN^-) = +y
(HCN) = -y
equilibrium:
(HCN) = 1.24 - y
(H^+) = 0 + y = y
(CN^-) = 0 + y = y
Substitute the equilibrium concns into the Ka expression and solve for y = (H^+), then pH = -log(H^+).
To solve this question, you need to use the concept of the acid dissociation constant (Ka) and the relationship between Ka and the pH of a solution.
Here is how you can find the pH of the solution:
Step 1: Write the balanced equation for the dissociation of HCN(aq):
HCN(aq) ⇌ H+(aq) + CN-(aq)
Step 2: Write the expression for the acid dissociation constant (Ka):
Ka = [H+(aq)][CN-(aq)] / [HCN(aq)]
Step 3: Since the concentration of HCN(aq) is given as 1.24 mol/L, we can assume that the concentration of [HCN(aq)] is equal to this value.
Step 4: Calculate the concentration of [H+(aq)] using the Ka value and the known concentration of [HCN(aq)]:
Ka = [H+(aq)][CN-(aq)] / [HCN(aq)]
6.2 x 10^-10 = [H+(aq)]^2 / 1.24
[H+(aq)]^2 = 6.2 x 10^-10 x 1.24
[H+(aq)]^2 = 7.688 x 10^-10
[H+(aq)] = √(7.688 x 10^-10)
[H+(aq)] ≈ 2.77 x 10^-5 mol/L
Step 5: Calculate the pH using the concentration of [H+(aq)]:
pH = -log[H+(aq)]
pH = -log(2.77 x 10^-5)
pH ≈ 4.56
Therefore, the pH of the 1.24 mol/L solution of HCN(aq) is approximately 4.56.