For a solution that is 0.20 M HCN(aq) with a Ka=4.9x10^-10
a. calculate the [H+]
b. calculate the [OH-]
c. what's the pH?
Please show your work on how to solve it. Thank you
Thanks! That was helpful.
thunder blew a 3-1 lead
To solve these questions, we need to use the equation for the dissociation of HCN, which is:
HCN (aq) ⇌ H+ (aq) + CN- (aq)
Given that the concentration of HCN is 0.20 M and the Ka value is 4.9x10^-10, let's solve each part step-by-step:
a. Calculate [H+]:
To find the concentration of [H+], we need to calculate the equilibrium concentration of H+ ions when HCN dissociates.
Let's assume x is the concentration of H+ ions formed. The equilibrium concentrations for HCN, H+, and CN- will then be (0.20 - x), x, and x, respectively.
Since the equilibrium constant (Ka) expression is given by:
Ka = [H+][CN-] / [HCN]
Substitute the expressions for each concentration into the expression:
Ka = x * x / (0.20 - x)
Since the value of x is expected to be small compared to 0.20, we can approximate (0.20 - x) as 0.20 for simplicity.
So, the equation becomes:
4.9x10^-10 = x^2 / 0.20
To solve for x, let's multiply both sides by 0.20:
(4.9x10^-10) * 0.20 = x^2
9.8x10^-11 = x^2
Taking the square root of both sides:
x ≈ √(9.8x10^-11)
x ≈ 3.13x10^-6
Therefore, the concentration of [H+] in the solution is approximately 3.13x10^-6 M.
b. Calculate [OH-]:
To calculate the concentration of [OH-], we can use the fact that the equation for water dissociation is:
H2O ⇌ H+ + OH-
Since the concentration of [H+] is known to be 3.13x10^-6 M, we know that the concentration of [OH-] will be the same since water is neutral. Therefore, [OH-] ≈ 3.13x10^-6 M.
c. Calculate the pH:
The pH is a measure of the acidity of a solution and is defined as the negative logarithm (base 10) of the concentration of [H+].
pH = -log[H+]
Substituting the value of [H+] that we calculated:
pH = -log(3.13x10^-6)
pH ≈ 5.50
Therefore, the pH of the solution is approximately 5.50.
.............HCN ==> H^+ + CN^-
initial....0.20M.....0.....0
change......-x.......+x.....+x
final.... 0.20-x......x.....x
Ka = ((H^+)(CN^-)/(HCN)
Substitute from the above and solve for (H^+).
pH = -log(H^+)
pH + pOH = pKw = 14
Post your work if you get stuck.