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Chapter 14: Problem 45

Carbonate buffers are important in regulating the pH of blood at \(7.40 .\) If the carbonic acid concentration in a sample of blood is 0.0012 \(M,\) determine the bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\). \(\mathrm{H}_{2} \mathrm{CQ}_{3}(a g) \rightleftharpoons \mathrm{HCQ}_{3}^{-}(a g)+\mathrm{H}^{+}(a g) \quad \) \(K_{\mathrm{a}}=4.3 \times 10^{-7}\)

Short Answer

Expert verified
To maintain the pH of blood at 7.40, the bicarbonate ion concentration (\(HCO_3^-\)) must be approximately 0.00234 M when the carbonic acid concentration is 0.0012 M.

Step by step solution

01

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH, pKa, acid concentration, and conjugate base concentration in a solution. It has the form: \[pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right)\] where: - \(pH\) is the pH of the solution - \(pKa\) is the negative logarithm of the acid dissociation constant, \(K_a\) - \([A^-]\) is the concentration of the conjugate base (bicarbonate ion, \(HCO_3^-\)) - \([HA]\) is the concentration of the weak acid (carbonic acid, \(H_2CO_3\)) We will use this equation to solve for the bicarbonate ion concentration (\([HCO_3^-]\)).
02

Calculate the pKa

In order to use the Henderson-Hasselbalch equation, we first need to find the pKa. The pKa is the negative logarithm of Ka, which is given as \(4.3 \times 10^{-7}\). \[pKa = -\log(K_a)\] Plug in the value of Ka: \[pKa = -\log(4.3 \times 10^{-7})\] Calculate the pKa value: \[pKa \approx 6.366\]
03

Plug known values into the Henderson-Hasselbalch equation

We can now update the Henderson-Hasselbalch equation with the known values of pH, pKa, and the carbonic acid concentration (\([H_2CO_3]\)): \[7.40 = 6.366 + \log\left(\frac{[HCO_3^-]}{0.0012}\right)\]
04

Solve for bicarbonate ion concentration ([HCO_3^-])

To find the bicarbonate ion concentration, first subtract the pKa value from the pH. \[7.40 - 6.366 = \log\left(\frac{[HCO_3^-]}{0.0012}\right)\] Next, remove the logarithm using the exponentiation with base 10. \[\frac{[HCO_3^-]}{0.0012} = 10^{(7.40 - 6.366)}\] Now, multiply both sides by 0.0012 to find the bicarbonate ion concentration. \[[HCO_3^-] = 0.0012 \times 10^{(7.40 - 6.366)}\] Calculate the value for [HCO_3^-]: \[[HCO_3^-] \approx 0.00234\,M\]
05

Interpret the result

To maintain the pH of blood at 7.40, the bicarbonate ion concentration (\(HCO_3^-\)) must be approximately 0.00234 M when the carbonic acid concentration is 0.0012 M.

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Most popular questions from this chapter

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between 7.1 and 7.2 a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

In the titration of \(50.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) methylamine, \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with \(0.50 M\) HCl, calculate the pH under the following conditions. a. after \(50.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl has been added b. at the stoichiometric point

Which of the following mixtures would result in a buffered solution when 1.0 L of each of the two solutions are mixed? a. \(0.2 M\) HNO and \(0.4 M \mathrm{NaNO}_{3}\) b. \(0.2 M \mathrm{HNO}_{3}\) and \(0.4 \mathrm{M} \mathrm{HF}\) c. \(0.2 \mathrm{M} \mathrm{HNO}_{3}\) and \(0.4 \mathrm{M} \mathrm{NaF}\) d. \(0.2 M\) HNO \(_{3}\) and \(0.4 M\) NaOH

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After \(23.75 \mathrm{mL}\) of the weak acid solution has been added to \(50.0 \mathrm{mL}\) of the \(0.100 \mathrm{M} \mathrm{NaOH}\) solution, the \(\mathrm{pH}\) of the resulting solution is \(10.50 .\) Calculate the original concentration of the solution of weak acid.

You have \(75.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HA. After adding \(30.0 \mathrm{mL}\) of \(0.10 M \mathrm{NaOH},\) the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA} ?\)
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