calculate the pH of a 0.50 m solution of HCN. Ka for HCN is 4.9*10^-10
(The chemical equation for this.)
The chemical equation for the dissociation of HCN in water is:
HCN (aq) ⇌ H+ (aq) + CN- (aq)
Now, we need to calculate the concentration of H+ ions in the solution, which will allow us to determine the pH.
First, we can determine the initial concentration of HCN (0.50 M) as it is a monoprotic acid. Since HCN is a weak acid, it does not completely dissociate in water.
Let's assume that "x" mol/L of HCN dissociates. Therefore, the concentration of H+ ions is also "x" mol/L.
Using the equation for the ionization of HCN, we can set up an expression for the equilibrium constant, Ka:
Ka = [H+][CN-]/[HCN]
We know that the initial concentration of HCN is 0.50 M, and the concentration of CN- (which comes from the dissociation of HCN) is also x mol/L.
Therefore, the equation becomes:
Ka = (x)(x)/(0.50 - x)
Substituting the value of Ka (4.9 × 10^-10), we can solve for "x" using the quadratic formula:
4.9 × 10^-10 = (x)(x)/(0.50 - x)
Rearranging the equation, we get:
x^2 = (4.9 × 10^-10)(0.50 - x)
x^2 = 2.45 × 10^-10 - (4.9 × 10^-10)x
x^2 + (4.9 × 10^-10)x - 2.45 × 10^-10 = 0
Using the quadratic formula, we can solve for "x":
x = [-b ± √(b^2 - 4ac)] / (2a)
Plugging in the values, we get:
x = [-(4.9 × 10^-10) ± √((4.9 × 10^-10)^2 - 4(1)(-2.45 × 10^-10))] / (2(1))
Simplifying this equation will give us the value of "x," which represents the concentration of H+ ions. Once we have "x," we can calculate the pH using the formula:
pH = -log[H+]
Let me do the calculations for you.
To calculate the pH of a solution of HCN (hydrogen cyanide), we need to use the dissociation constant (Ka) and the initial concentration of the acid.
The chemical equation for the dissociation of HCN is as follows:
HCN ⇌ H+ + CN-
Since HCN is a weak acid, it only partially dissociates in water. The equilibrium expression for its dissociation is:
Ka = [H+][CN-] / [HCN]
Given that the Ka value for HCN is 4.9 x 10^-10 and the initial concentration of HCN is 0.50 M, we can use this information to calculate the concentration of H+ (hydrogen ions) in the solution and subsequently determine the pH.
Let's assume that x represents the concentration of H+ ions formed upon the dissociation of HCN. As HCN dissociates, x mol/L of H+ and CN- ions are formed, while the remaining concentration of HCN will be (0.50 - x) M.
Using the equilibrium expression, we can write:
(4.9 x 10^-10) = (x)(x) / (0.50 - x)
Now, solve this equation to find the value of x, which represents the concentration of H+ ions. This can be done using algebraic manipulation or by using an online calculator or software.
Once you have the value of x, you can calculate the pH using the equation:
pH = -log10[H+]
Just substitute the concentration of H+ ions into the equation to find the pH.
Got got that concept.
Found that x=1.56 x10^-5 (x is H+)
PH=-log(H+)
=-log(1.56x10^-5)
PH= 4.80. (Is that correct)
If you mean m it is a tougher problem. I see you can't seem to capitalize the first letter of a sentence so you PROBABLY M which stands for molarity. (m doesn't, it stands for molality.)
.........HCN ==> H^+ + CN^-
I........0.50....0.......0
C........-x......x.......x
E.......0.50-x...x.......x
Substitute the E line into the Ka expression and solve for H^+, then convert to pH.