Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10.0 mL of buffer of each of the following pH values?
a. Ph 3.7
b. pH 5.7
C. pH 5.7
I just need to see how to do one of them and I'll understand how to do the rest.
pH = pka + log [base CH3COONa]/[acid CH3COOH]
3.7 = 4.7 + log [base]/[acid]
alegbra to remove 4.7 from both sides
-1.0 = log [base]/[acid]
log= 10^x so alegbra to remove 10^x
10^-1 = [base]/[acid] or 0.10 = [base]/[acid]
1 ml for base added to 9 ml of acid
2.4
To determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare a buffer of a specific pH value, you will need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pH is the desired pH value, pKa is the acid dissociation constant of the weak acid component (CH3COOH) of the buffer, [A-] is the concentration of the conjugate base component (CH3COO-) of the buffer, and [HA] is the concentration of the weak acid component (CH3COOH) of the buffer.
Let's work through an example for pH 3.7:
a. pH 3.7
The pKa value for acetic acid (CH3COOH) is 4.74. From the Henderson-Hasselbalch equation, we have:
3.7 = 4.74 + log([A-]/[HA])
To solve for the ratio [A-]/[HA], we can rearrange the equation:
log([A-]/[HA]) = 3.7 - 4.74
[A-]/[HA] = 10^(3.7 - 4.74)
Now, we need to choose an appropriate ratio of [A-]/[HA]. Typically, a buffer is most effective when the ratio is close to 1. Therefore, we can assume an equal concentration of [A-] and [HA]:
[A-] = [HA]
Substituting this assumption into the equation, we have:
[A-]/[HA] = 10^(3.7 - 4.74)
[A-]^2 = 10^(3.7 - 4.74)
Taking the square root of both sides, we can solve for [A-]:
[A-] = sqrt(10^(3.7 - 4.74))
Now that we have the concentration of the conjugate base component ([A-]) of the buffer, we can calculate the volume of each component required.
For a total volume of 10.0 mL, let's assume the volume of CH3COOH is x mL, which means the volume of CH3COONa will be (10 - x) mL.
Using the molarity (0.10 M) of both CH3COOH and CH3COONa, we can set up the equation:
x(0.10 M) = [CH3COOH]
(10 - x)(0.10 M) = [CH3COONa]
Substituting the calculated concentrations:
x(0.10 M) = sqrt(10^(3.7 - 4.74))
(10 - x)(0.10 M) = sqrt(10^(3.7 - 4.74))
Now, you can solve these equations simultaneously to determine the volumes of CH3COOH and CH3COONa required to prepare the buffer at pH 3.7.
1.5
Plug and chug into the HH equation.
These problelms, where no concentration is given for the buffer, is a wide open problem with no restrictions. Here is what you do.
Let x volume(in mL) CH3COONa and 10-x = volume in mL of CH3COOH.
pKa CH3COOH is about 4.74 but you need to use the value in your text.
3.7 = 4.74 + log (0.1x)/0.1(10-x)
Solve for x ad 10-x