Find the pH of a buffer that consists of 0.82 M methylamin (CH3NH2) and 0.71 M CH3NH3Cl (pKb of methylamine (CH3NH2) = 3.35.)
I keep getting 3.41, but it says that it's wrong...
CH3NH2=0.71M=B
CH3NH3Cl=0.82M=BH
B+H2O--->BH +OH
kb=[OH][BH]/{B}
kb=[OH]*([BH]/[B])
-logkb=-log[OH]*-log([BH]/[B])
pkb=pOH-log([BH]/[B])
pOH=pKb+log([BH]/[B]
pH=14-pOH
Essentially, I was thinking this problem out by deriving the Henderson-Hasselbalch equation. You only need the following information to calculate your answer:
CH3NH2=0.71M=B
CH3NH3Cl=0.82M=BH
pOH=pKb+log([BH]/[B]
pH=14-pOH
To find the pH of a buffer solution, we need to determine the concentration of the common ion, in this case, CH3NH3+. The pKb is given as 3.35, which means we can use it to find the pKa of CH3NH2, using the following equation:
pKw = pKa + pKb
where pKw is the ion product of water and is approximately equal to 14.
14 = pKa + 3.35
pKa = 14 - 3.35
pKa = 10.65
Now that we have the pKa, we can calculate the concentration of CH3NH3+ using the Henderson-Hasselbalch equation:
pH = pKa + log(CH3NH3+/CH3NH2)
pH = 10.65 + log(CH3NH3+/0.82)
To find the concentration of CH3NH3+, we can use the initial concentration of CH3NH3Cl and the dissociation equilibrium:
CH3NH3+ + Cl- ⇌ CH3NH2 + HCl
Let's assume the concentration of CH3NH3+ is x. Since CH3NH3+ and Cl- dissociate in a 1:1 ratio, the concentration of CH3NH2 will be the initial concentration minus x, and the concentration of Cl- will also be x.
So, the equilibrium expression for the dissociation of CH3NH3Cl is:
Kc = [CH3NH2][HCl]/[CH3NH3+][Cl-]
The initial concentration of CH3NH3Cl is 0.71 M, and at equilibrium, the concentration of CH3NH3+ is x, and the concentration of CH3NH2 is 0.82 - x.
Using the given pKb of methylamine, we can calculate the value of Kc:
Kc = 10^(-pKb)
Kc = 10^(-3.35)
Now, plug these values into the equilibrium expression:
10^(-3.35) = (0.82 - x)(x)/(0.71)(x)
Simplifying further:
10^(-3.35) = (0.82 - x)/0.71
Solve this equation for x. Once you find the value of x, substitute it into the pH equation:
pH = 10.65 + log(x/0.82)
This will give you the correct pH of the buffer solution.
To find the pH of a buffer, we need to consider the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa or pKb of the conjugate acid-base pair and the ratio of their concentrations.
The Henderson-Hasselbalch equation for acids is:
pH = pKa + log([A-]/[HA])
The Henderson-Hasselbalch equation for bases is:
pOH = pKb + log([BH+]/[B])
In this case, we have a base, methylamine (CH3NH2), and its conjugate acid, methylammonium chloride (CH3NH3Cl). The pKb of methylamine is 3.35.
To find the pH, we need to use the Henderson-Hasselbalch equation for bases because we have a base and its conjugate acid. We can calculate the pOH and then convert it to pH using the equation: pH = 14 - pOH.
Let's start by calculating the pOH:
pOH = pKb + log([BH+]/[B])
The base is methylamine (CH3NH2) and its conjugate acid is methylammonium chloride (CH3NH3Cl). The concentration of methylamine ([B]) is 0.82 M, and the concentration of methylammonium chloride ([BH+]) is 0.71 M.
Substituting these values into the equation, we get:
pOH = 3.35 + log(0.71/0.82)
Now let's calculate the pOH:
pOH = 3.35 + log(0.8659)
pOH ≈ 3.35 - 0.0623
pOH ≈ 3.29
Finally, we can calculate the pH using the equation:
pH = 14 - pOH
pH = 14 - 3.29
pH ≈ 10.71
Therefore, the pH of the buffer solution is approximately 10.71.