Determine pH of each solution:
b) 0.20 M CH3NH3I
I did....
CH3NH3^+ +H2O -> CH3NH2 + H3O^+
0.20 ---------------0--------------0
0.20-x--------------x------------x
so i got...
? =x^2/0.20-x
i looked up in the back of the book and found the Kb of CH3NH2 to be 4.4 X 10^-4.
do i put this number in for the question mark up above or do i have to do that Ka=Kw/Kb thing? How do i know if i use ka or kb?....
Well, determining the pH of a solution can be tricky, but don't worry, I've got a joke to lighten things up!
Why don't scientists trust atoms?
Because they make up everything!
Okay, back to your question. In this case, you're dealing with the dissociation of CH3NH3I, which is a salt formed from a weak base, CH3NH2, and a strong acid, HI. Since HI is a strong acid, it will completely dissociate in water, so we can focus on the CH3NH2 component.
To determine the pH, we need to find the concentration of H3O+ ions, which is equivalent to the concentration of CH3NH2 that has dissociated. Since CH3NH2 is a weak base, we can use the Kb (base dissociation constant) to find the concentration of OH- ions and then convert that to H3O+ concentration using Kw (the water dissociation constant).
So, to answer your question, you'll need to use the Kb value of CH3NH2 (4.4 x 10^-4) to find the concentration of OH- ions and then convert that to H3O+ concentration using Kw.
To determine the pH of the solution, you need to use the Kb value because CH3NH2 is a weak base. The Kb value represents the equilibrium constant for the reaction between CH3NH2 and water.
To calculate the pH of the solution, you can use the Kb value and the equilibrium expression for the reaction:
Kb = [CH3NH2][H3O+]/[CH3NH3+]
Since the concentration of CH3NH3+ is known as 0.20 M, you can substitute the values into the equation:
4.4 × 10^-4 = [CH3NH2][H3O+]/0.20
Assuming x is the concentration of CH3NH2 and H3O+, you can write the expression as:
4.4 × 10^-4 = (x)(x)/0.20
Simplifying the equation further:
4.4 × 10^-4 = x^2/0.20
Now, you need to solve this quadratic equation to find the concentration of H3O+ (which can be used to calculate the pH).
To do this, you can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 0, and c = -4.4 × 10^-4. Plugging in these values:
x = (0 ± √(0 - 4(1)(-4.4 × 10^-4))) / (2(1))
x = √(4.4 × 10^-4) / 2
x = (2 × 10^-2) / 2
x = 10^-2
Now you have the concentration of H3O+, which is 10^-2 M. To calculate the pH, you can use the formula:
pH = -log[H3O+]
pH = -log(10^-2)
pH = -(-2)
pH = 2
Therefore, the pH of the 0.20 M CH3NH3I solution is 2.
To determine the pH of the solution, you have correctly started by writing the balanced chemical equation for the reaction of CH3NH3I with water. However, there seems to be a small mistake in the equation you wrote. The correct equation for the reaction is:
CH3NH3I + H2O -> CH3NH3OH + HI
Now let's assign variables to the initial concentration and changes for each species in the equation:
Initial concentration of CH3NH3I = 0.20 M (let's call this [CH3NH3I])
Change in CH3NH3I = -x
Change in CH3NH3OH = x
Change in HI = x
Writing the expression for the equilibrium constant (Kb) for the reaction:
Kb = ([CH3NH3OH][HI])/[CH3NH3I]
As you mentioned, the Kb value for CH3NH2 (the conjugate base of CH3NH3OH) is 4.4 x 10^-4. Therefore, you will use this value to solve for x.
Since the concentration of CH3NH3I is 0.20 M, it can be assumed that x is insignificant compared to 0.20 M. This allows us to ignore the change (x) to the original concentration.
Now, substitute the known values into the equation:
4.4 x 10^-4 = (x)(x)/(0.20)
Simplifying:
4.4 x 10^-4 = x^2/0.20
Now, multiply both sides of the equation by 0.20:
0.20 × 4.4 x 10^-4 = x^2
8.8 x 10^-5 = x^2
Taking the square root of both sides:
x = √(8.8 x 10^-5)
x ≈ 0.00937 M
Now, you have found the concentration of CH3NH3OH (which is equal to [H3O^+]) to be approximately 0.00937 M. To convert this to pH, use the equation:
pH = -log[H3O^+]
pH = -log(0.00937)
pH ≈ 2.03
Therefore, the pH of the solution containing 0.20 M CH3NH3I is approximately 2.03.
You looked up Kb which is right.
You want Ka for CH3NH3^+ and they don't make tables for those.
You ALWAYS know that KaKb=Kw; therefore, if you have Kb and want Ka it's Ka = Kw/Kb. If you have Ka and want Kb it's Kb = Kw/Ka.