An aqueous solution contains 0.445 M ethylamine (C2H5NH2). How many mL of 0.226 M hydrochloric acid would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 10.200
When you post a problem like this you should tell us what you are using for pKa or pKb; otherwise we can come up with the same answers you may have since tables containing those pK values are not consistent from text to text.
You have 0.445 x 150 mL = approx 67 millimols of the base that that's an estimate and you should do it more carefully than that.
.......CH3NH2 + HCl ==> CH3NH3^+ + Cl^-
I.......67...............0
add..............x..............
C........-x.....-x.......x
E......67-x......0........x
Now substitute the E line into the Henderson-Hasselbalch equation and solve for x = millimols HCl to be added.
Then mmols HCl = M x mL. You know mmols and M, solve for mL.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer to the concentration of the acidic and basic components.
The Henderson-Hasselbalch equation is:
pH = pKa + log([A-]/[HA])
Where:
pH is the desired pH of the buffer
pKa is the acid dissociation constant of the weak acid (ethylamine in this case)
[A-] is the concentration of the conjugate base (ethylamine in this case)
[HA] is the concentration of the weak acid (ethylammonium ion, C2H5NH3+, in this case)
First, we need to determine the pKa of ethylamine. The pKa is the negative logarithm of the acid dissociation constant (Ka). The dissociation constant (Ka) for ethylamine is 6.4 * 10^-4. Therefore, the pKa is calculated as follows:
pKa = -log(6.4 * 10^-4) = 3.19
Next, we need to determine the concentrations of ethylamine ([A-]) and ethylammonium ion ([HA]) in the solution.
Given that the concentration of ethylamine is 0.445 M, and assuming complete dissociation, the concentration of ethylammonium ion would also be 0.445 M.
Now, let's calculate the ratio of [A-] to [HA]:
[A-]/[HA] = 0.445 M / 0.445 M = 1
Substituting the known values into the Henderson-Hasselbalch equation:
10.200 = 3.19 + log(1)
Since log(1) is equal to 0, we can simplify the equation to:
10.200 = 3.19
However, this equation is not true, which means it is not possible to prepare a buffer with a pH of 10.200 using the given concentrations of ethylamine and ethylammonium ion.
Therefore, it is not possible to answer the question regarding the volume of 0.226 M hydrochloric acid needed to prepare a buffer with a pH of 10.200 using the given concentrations.