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Chapter 24: Problem 102

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

Expert verified
The ratio of [AMP-OH⁻] to [AMP-O²⁻] in blood at pH 7.4 is approximately 1:1.54.

Step by step solution

01

Write down the Henderson-Hasselbalch equation.

The equation we will use is: pH = pKa + log([A⁻]/[HA]) Where pH is the blood pH, pKa is the pKa of the monoanion AMP, [A⁻] is the concentration of AMP-O²⁻, and [HA] is the concentration of AMP-OH⁻.
02

Plug in the given values.

We are given the following values: pH = 7.4 pKa = 7.21 We will now plug these values into the Henderson-Hasselbalch equation: 7.4 = 7.21 + log([AMP-O²⁻]/[AMP-OH⁻])
03

Solve for the ratio of [AMP-OH⁻] to [AMP-O²⁻].

Rearrange the equation to isolate the concentration ratio on one side: log([AMP-O²⁻]/[AMP-OH⁻]) = 7.4 - 7.21 Now, calculate the difference: log([AMP-O²⁻]/[AMP-OH⁻]) = 0.19 To remove the logarithm, use the inverse function, which is the exponential function: [AMP-O²⁻]/[AMP-OH⁻] = 10^0.19 Calculate the exponent: [AMP-O²⁻]/[AMP-OH⁻] ≈ 1.54
04

Write down the final answer.

The ratio of [AMP-OH⁻] to [AMP-O²⁻] in blood at pH 7.4 is approximately 1:1.54.

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