Answer:
Buffer A: pH 5.25; 2.82 mol of acetate for every 1 mol of acetic acid
Buffer B: pH 9.25; 11.2 mol of bicarbonate for every 1 mol of carbonate
Explanation:
The pI of a protein is the pH at which it has a net charge of zero.
The protein has a positive charge below the pI and a negative charge above it.
Since pI = 7.25, let's make one buffer at pH = 5.25 and one at pH 9.25.
1. Preparation of pH 5.25 buffer
(a) Choose the conjugate pair
We should choose a buffer with pKₐ close to 5.25.
Acetic acid has pKₐ = 4.8, so let's make acetate buffer.
(b) Protocol for preparation
We can use the Henderson-Hasselbalch equation to get the acid/base ratio.
![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\5.25& = & 4.8 +\log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\0.45& = & \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\2.82 & = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\end{array}\\\text{The base/acid ratio must be $\mathbf{2.82:1}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Ctext%7BpH%7D%20%26%20%3D%20%26%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C5.25%26%20%3D%20%26%204.8%20%2B%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C0.45%26%20%3D%20%26%20%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C2.82%20%26%20%3D%20%26%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20base%2Facid%20ratio%20must%20be%20%24%5Cmathbf%7B2.82%3A1%7D%24%7D)
To make the buffer, mix the solutions to get a ratio of 2.82 mol of acetate for every 1 mol of acetic acid.
2. Preparation of pH 9.25 buffer
(a) Choose the conjugate pair
We should choose a buffer with pKₐ close to 9.25.
Bicarbonate has pKₐ = 10.3, so let's make a pH 9.25 bicarbonate buffer.
(b) Protocol for preparation
![\begin{array}{rcl}\text{pH} & = & \text{pK}_{\text{a}} + \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\9.25& = & 10.3 +\log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\-1.05& = & \log \left(\dfrac{[\text{A}^{-}]}{\text{[HA]}}\right )\\\\0.08913& = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\dfrac{1}{11.2}& = &\dfrac{[\text{A}^{-}]}{\text{[HA]}}\\\\\end{array}\\\text{The acid/base ratio must be $\mathbf{11.2:1}$}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Ctext%7BpH%7D%20%26%20%3D%20%26%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C9.25%26%20%3D%20%26%2010.3%20%2B%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C-1.05%26%20%3D%20%26%20%5Clog%20%5Cleft%28%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5Cright%20%29%5C%5C%5C%5C0.08913%26%20%3D%20%26%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Cdfrac%7B1%7D%7B11.2%7D%26%20%3D%20%26%5Cdfrac%7B%5B%5Ctext%7BA%7D%5E%7B-%7D%5D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20acid%2Fbase%20ratio%20must%20be%20%24%5Cmathbf%7B11.2%3A1%7D%24%7D)
To make the buffer, mix the solutions to get a ratio of 11.2 mol of bicarbonate for every 1 mol of carbonate.