Question

The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism: where \(A=\) adenosine. If the \(\mathrm{p} K_{a}\) for this anion is 7.21 what is the ratio of \(\left[\mathrm{AMP}-\mathrm{OH}^{-}\right]\) to \(\left[\mathrm{AMP}-\mathrm{O}^{2-}\right]\) in blood at \(\mathrm{pH} 7.4\) ?

Short answer

The ratio of \([\mathrm{AMP}-\mathrm{OH}^{-}]\) to \([\mathrm{AMP}-\mathrm{O}^{2-}]\) in blood at pH 7.4 is approximately 1.54.

Step-by-step solution

Remember the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation helps us relate the pKa of a weak acid to its pH and the ratio of conjugate base to the acid: \[ pH = pKa + \log\frac{[\text{A}^-]}{[\text{HA}]} \] In this case, A^- represents AMP-O^2-, and HA represents AMP-OH^-. We need to find the ratio of \([\mathrm{AMP}-\mathrm{OH}^{-}]\) to \([\mathrm{AMP}-\mathrm{O}^{2-}]\), which corresponds to the \(\frac{[\text{A}^-]}{[\text{HA}]}\) ratio in the equation.

Plug in the known values

We know the pH of the blood (7.4) and the pKa for AMP's monoanion (7.21). We can now plug these values into the Henderson-Hasselbalch equation: \[ 7.4 = 7.21 + \log\frac{[\text{AMP-O}^{2-}]}{[\text{AMP-OH}^{-}]} \]

Solve for the ratio

Now we need to isolate the ratio \(\frac{[\text{AMP-O}^{2-}]}{[\text{AMP-OH}^{-}]}\) and find its value: \[ 7.4 - 7.21 = \log\frac{[\text{AMP-O}^{2-}]}{[\text{AMP-OH}^{-}]} \] \[ 0.19 = \log\frac{[\text{AMP-O}^{2-}]}{[\text{AMP-OH}^{-}]} \] To calculate the ratio \(\frac{[\text{AMP-O}^{2-}]}{[\text{AMP-OH}^{-}]}\), we need to take the antilog of 0.19: \[ \frac{[\text{AMP-O}^{2-}]}{[\text{AMP-OH}^{-}]} = 10^{0.19} \]

Find the final ratio

Now calculate the numerical value for the antilog of 0.19: \[ \frac{[\text{AMP-O}^{2-}]}{[\text{AMP-OH}^{-}]} = 10^{0.19} \approx 1.54 \] Therefore, the ratio of \([\mathrm{AMP}-\mathrm{OH}^{-}]\) to \([\mathrm{AMP}-\mathrm{O}^{2-}]\) in blood at pH 7.4 is approximately 1.54.

Key concepts

Understanding the pKa and pH Relationship

The relationship between pKa and pH is critical to understanding how acids and bases behave in various environments, such as in blood or other biological fluids. The pKa represents the acidity constant, which is a quantitative measure of the strength of an acid in solution. It specifically indicates the pH value at which half of the molecules of the weak acid lose a proton.

In other words, at the pKa, the concentrations of the acid (HA) and its conjugate base (A-) are equal. When the system's pH is lower than the pKa, the acid form (HA) predominates because the environment is more acidic, encouraging the acid to retain its proton. Conversely, when the pH is higher than the pKa, the environment is more basic, thus favoring the form of the conjugate base (A-), which has given up its proton.

The Henderson-Hasselbalch equation is a formula derived from the acid dissociation constant that mathematically establishes this relationship. It helps us determine the ratio of the protonated form to the deprotonated form of a substance at any given pH. Through it, you can predict the charge state of the acid-base pair, which is crucial for understanding many biological processes and drug action mechanisms.

Conjugate Acid-Base Pairs

Conjugate acid-base pairs are two substances that transform into each other by the gain or loss of a proton. In the context of the Henderson-Hasselbalch equation, the acid (HA) and its conjugate base (A-) form such a pair.

The important part to understand about conjugate acid-base pairs is that they are in equilibrium. The behavior of these pairs is subject to the principles of acid-base equilibrium. When an acid donates a proton, it becomes its conjugate base, and when a base gains a proton, it turns into its conjugate acid. This proton transfer equilibrium reflects the reversible nature of acid-base reactions.

In our exercise, the AMP-OH⁻ acts as the acid, and AMP-O₂⁻ is the associated conjugate base. When AMP-OH⁻ loses a proton (H⁻), it turns into AMP-O₂⁻. The ratio of these two forms at a particular pH value can be calculated using the Henderson-Hasselbalch equation, revealing how the protonation state shifts with changing pH levels.

Understanding Acid-Base Equilibrium

Acid-base equilibrium is the balance that occurs in a solution between the forward and reverse reactions of acid and base, along with their respective conjugate base and acid. It's a dynamic state where the rate of the forward reaction (acid losing a proton) is equal to the rate of the reverse reaction (conjugate base gaining a proton).

The state of acid-base equilibrium is essential to maintaining bodily functions and cellular activities within a narrow pH range. Enzymes, which work optimally at certain pH levels, rely on this balance, as do various metabolic pathways.

In the exercise with AMP, an intermediate in phosphate metabolism, we analyzed this equilibrium in the context of a biological system - the blood - which typically has a pH around 7.4. In either absorbing or donating protons, the substances involved in acid-base reactions act to buffer changes in pH, preserving optimal conditions for biological reactions.