What concentration of CH3NH3Br is necessary to prepare a pH = 10.00 buffer solution assuming the base concentration is 0.49 M (Kb for CH3NH2 = 4.4 x 10-4).
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To prepare a buffer solution with a pH of 10.00, we need a solution containing a weak base, CH3NH2 (methylamine), and its conjugate acid, CH3NH3+ (methylammonium ion).
First, let's set up the balanced equation for the reaction between CH3NH2 and water:
CH3NH2 + H2O ⇌ CH3NH3+ + OH-
The equilibrium constant expression for this reaction is the base dissociation constant, Kb:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Given that the Kb for CH3NH2 is 4.4 x 10^-4, we can use this value to calculate the concentration of OH- in the solution at pH 10.00.
Using the equation for pOH:
pOH = -log10[OH-]
pOH = 14.00 - pH
pOH = 14.00 - 10.00
pOH = 4.00
Since pOH = -log10[OH-], we can convert pOH to [OH-]:
[OH-] = 10^(-pOH)
[OH-] = 10^(-4.00)
[OH-] = 0.0001 M
Now, to calculate the concentration of CH3NH2 needed for the buffer solution.
Since the concentration of OH- formed by the reaction between CH3NH2 and water is equal to the concentration of [CH3NH3+] in the buffer solution, we can use the equation for the Kb expression:
Kb = [CH3NH3+][OH-] / [CH3NH2]
Rearranging the equation, we have:
[CH3NH3+] = (Kb * [CH3NH2]) / [OH-]
[CH3NH3+] = (4.4 x 10^-4 * 0.49 M) / 0.0001 M
[CH3NH3+] = 2.156 M
Therefore, a concentration of 2.156 M of CH3NH3Br is necessary to prepare a pH = 10.00 buffer solution, assuming the base concentration is 0.49 M.
To determine the concentration of CH3NH3Br needed to prepare a pH = 10.00 buffer solution, we need to consider the Henderson-Hasselbalch equation for a buffer solution:
pH = pKa + log([A-]/[HA])
In this case, CH3NH3Br acts as the acidic component (HA) and CH3NH2 as the basic component (A-). The equation can be rearranged as follows:
[HA] = [A-] x 10^(pH - pKa)
To find the concentration of CH3NH3Br ([HA]), we first need to calculate the pKa of CH3NH3Br. Since CH3NH3Br is a salt formed by the reaction of CH3NH2 (methylamine) with HBr (hydrobromic acid), the pKa of CH3NH3Br can be calculated using the pKa = pKa (acid) + pKa (base) equation.
The pKa of CH3NH2 can be found using the Kb value:
Kw = Ka x Kb
Where Kw is the ion product of water (1 x 10^-14) and Ka is the acid dissociation constant of CH3NH3+. Rearranging the equation gives:
Ka = Kw / Kb
Substituting the values:
Ka = (1 x 10^-14) / (4.4 x 10^-4)
= 2.27 x 10^-11
Taking the logarithm of Ka gives:
log(Ka) = log(2.27 x 10^-11)
= -10.64
Therefore, the pKa of CH3NH2 is approximately 10.64.
Substituting the values into the Henderson-Hasselbalch equation and solving for [HA]:
pH = pKa + log([A-]/[HA])
10.00 = 10.64 + log([A-]/[HA])
We know that the base concentration is 0.49 M. Let's assume the concentration of CH3NH3Br is represented by x. Therefore, the concentration of CH3NH2 ([A-]) would be (0.49 - x).
Substituting these values into the equation:
10.00 = 10.64 + log((0.49 - x)/x)
To solve for x, we rearrange the equation:
log((0.49 - x)/x) = 10.00 - 10.64
(0.49 - x)/x = 10^(10.00 - 10.64)
Now we can calculate (0.49 - x)/x:
(0.49 - x)/x = 10^(-0.64)
0.49 - x = x x 10^(-0.64)
0.49 = x(1 + 10^(-0.64))
x = 0.49 / (1 + 10^(-0.64))
By calculating this expression, you will find the concentration of CH3NH3Br needed to prepare the pH = 10.00 buffer solution assuming the base concentration is 0.49 M.