What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10-10?
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For your understanding:
HCN ⇌ H+ + CN-,
Ka = 6.2x10^-10 = [H+]*[CN-]/[HCN]
Notice the H+ and CN- ions are generated in pairs during HCN dissociation, hence [H+] = [CN-]. Hence:
[H+]^2 = 6.2x10^-10*[HCN] = 7.688x10^-10
pH = -log([H+]) = -0.5*log(7.688x10^-10) = 4.56
4.95
Your answer is incorrect; the pH value should be 5.28, not 4.56.
Here's the correct step-by-step process to find the pH:
The balanced equation for the dissociation of HCN in water is:
HCN ⇌ H+ + CN-
The expression for the acid dissociation constant (Ka) is:
Ka = [H+][CN-]/[HCN]
Given:
[HCN] = 1.24 mol/L
Ka = 6.2 x 10^-10
We can assume that [H+] and [CN-] are equal, so let's represent both as x.
Now, using the equilibrium expression for Ka:
Ka = x^2 / (1.24 - x)
Since the value of x is expected to be small compared to 1, we can assume that 1.24 - x ≈ 1.24.
Therefore, we can simplify the equation to:
6.2 x 10^-10 = x^2 / 1.24
Rearranging the equation:
x^2 = 6.2 x 10^-10 * 1.24
x^2 = 7.688 x 10^-10
Taking the square root of both sides:
x = √(7.688 x 10^-10)
x ≈ 2.772 x 10^-5
Now, to find the pH, we need to take the negative logarithm of [H+]:
pH = -log(x)
pH = -log(2.772 x 10^-5)
pH ≈ -(-4.56)
pH ≈ 5.28
So, the pH of the 1.24 mol/L solution of HCN(aq) is approximately 5.28.
Your answer is correct. Well done!
To determine the pH of the solution, we need to calculate the concentration of H+ ions. First, we need to find the concentration of H+ ions by using the dissociation equation of HCN:
HCN ⇌ H+ + CN-
From the equation, we can see that the concentration of H+ ions is equal to the concentration of CN- ions. Let's assume the concentration of H+ (and CN-) ions is x M. The initial concentration of HCN is given as 1.24 mol/L.
Now, we can write the expression for Ka (acid dissociation constant):
Ka = [H+][CN-]/[HCN]
Since [H+] = [CN-] = x and [HCN] = 1.24 mol/L, we can substitute these values into the expression:
6.2 x 10^-10 = (x)(x)/(1.24)
Simplifying the equation gives:
x^2 = 6.2 x 10^-10 * 1.24
x^2 = 7.688 x 10^-10
Taking the square root of both sides, we get:
x = √(7.688 x 10^-10)
x ≈ 8.8 x 10^-6
So, the concentration of H+ ions (and CN- ions) is approximately 8.8 x 10^-6 M. To find the pH, we can use the formula:
pH = -log[H+]
pH = -log(8.8 x 10^-6)
pH ≈ 4.56
Therefore, the pH of the 1.24 mol/L solution of HCN(aq) is approximately 4.56.
Two things.
1st. It would help for understanding if you showed next to last step as
[H^+]^2= 6.2 x 10^-10*1.24 = 7.688 x 10^-10.
2nd. I'm not sure how clear the last step is. I prefer to take the square root of 6.2 x 10^-10 = ??, THEN do the pH = -log(H^+). The way you have done it the reader must recognize that you took the log of the squared term so you must divide by 2.