what is the pH of 0.80 M NaCN?
what is the pH of 0.80 M NaCN? What is the concentration of HCN in the solution?
I did this so far:
I 0.8..... 0....0
C -x.......x....x
E 0.8-x....x....x
(x)(x)/0.8-x
How do I solve for x?
You just need to set what you have equal to Kb for CN^- which isn't in any table but can be calculated this way.
Kb for CN^- = (Kw/Ka for HCN)
so (Kw/Ka for HCN) = (x)(x)/(0.8-x)
and solve for x.
what is the ph of 0.15 m NaCN
To solve for x, you can use the quadratic equation when dealing with weak acids or bases like HCN. The equilibrium constant expression for the dissociation of HCN is given by:
K_a = [H^+][CN^-] / [HCN]
In this case, we know that the initial concentration of NaCN is 0.80 M, and assuming complete dissociation, it will yield an equal concentration of cyanide ions (CN^-).
So, [CN^-] = 0.80 M.
Since HCN partially dissociates, the concentration of HCN is given by:
[HCN] = 0.80 - x
Using the equilibrium constant expression, we have:
K_a = [H^+][CN^-] / [HCN]
Given that Ka for HCN is 4.9 × 10^-10, we can substitute the known values into the equation:
4.9 × 10^-10 = x * 0.80 / (0.80 - x)
Now, let's solve for x. Let's multiply both sides of the equation by (0.80 - x):
4.9 × 10^-10 * (0.80 - x) = x * 0.80
3.92 × 10^-10 - 4.9 × 10^-10x = 0.80x
Collect the x terms on one side:
0.80x + 4.9 × 10^-10x = 3.92 × 10^-10
Combine like terms:
(0.80 + 4.9 × 10^-10)x = 3.92 × 10^-10
Divide both sides by (0.80 + 4.9 × 10^-10):
x = (3.92 × 10^-10) / (0.80 + 4.9 × 10^-10)
Now, plug in the values and solve for x using a calculator to get the concentration of HCN in the solution.
To solve for x, we first need to set up an equation based on the dissociation of NaCN in water. The general equation for this reaction is NaCN ⇌ Na+ + CN-. Since the CN- ion is the conjugate base of a weak acid (HCN), we can use the Ka expression for a weak acid to relate the concentration of HCN to the concentration of CN-.
The Ka expression for HCN is: Ka = [H+][CN-]/[HCN]
Since we are trying to find the concentration of HCN, we rearrange the equation to solve for [HCN]: [HCN] = [H+][CN-]/Ka
At equilibrium, the concentration of [H+] equals the concentration of [CN-]. Therefore, we can replace [H+] and [CN-] with x. So the equation becomes: [HCN] = x^2 / Ka
In this case, Ka is the acid dissociation constant of HCN, which is approximately 4.9 x 10^-10 at 25°C.
To find the value of x, we can substitute the given concentration of NaCN (0.80 M) into the equation for [HCN], as follows:
0.80 = x^2 / (4.9 x 10^-10)
Now, we can rearrange this equation to solve for x. Multiply both sides of the equation by (4.9 x 10^-10):
0.80 * (4.9 x 10^-10) = x^2
Simplify the left side of the equation:
3.92 x 10^-10 = x^2
To find the square root of both sides, take the square root of 3.92 x 10^-10:
x = √(3.92 x 10^-10)
Using a calculator, the value of x is approximately 6.25 x 10^-6.
Therefore, the concentration of HCN in the solution is approximately 6.25 x 10^-6 M.
To find the pH of the solution, you can use the formula pH = -log[H+], where [H+] is the concentration of hydrogen ions. Since [H+] is equal to the concentration of HCN, the pH of the solution would be -log(6.25 x 10^-6). Calculate this using a calculator to find the pH value.