Carbonate buffers are important in regulating the \(\mathrm{pH}\) 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 \(\mathrm{pH}=7.40\). \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7}\)

Short Answer

Expert verified
The bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\) is \(0.0013 M\).

Step by step solution

01

Equilibrium constant expression

Write the \(K_{a}\) expression using the given equilibrium constant: \[K_{a} = \frac{[\mathrm{HCO}_{3}^{-}][\mathrm{H}^{+}]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]}\]
02

Calculate the concentration of \(\mathrm{H}^{+}\) from the given \(\mathrm{pH}\)

To calculate the concentration of \(\mathrm{H}^{+}\) ions, use the \(\mathrm{pH}\) equation: \[\mathrm{pH} = -\log[\mathrm{H}^{+}]\] We are given \(\mathrm{pH}=7.40\), so solve for \([\mathrm{H}^{+}]\): \[7.40 = -\log[\mathrm{H}^{+}]\] \[[\mathrm{H}^{+}] = 10^{-7.40} = 3.98\times 10^{-8} M\]
03

Plug in to \(K_{a}\) expression and solve for \([\mathrm{HCO}_{3}^{-}]\)

Substitute the values from the given equilibrium constant, carbonic acid concentration, and the calculated \(\mathrm{H}^{+}\) concentration into the \(K_{a}\) expression: \[\begin{aligned} K_{a} &= \frac{[\mathrm{HCO}_{3}^{-}][\mathrm{H}^{+}]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]} \\ 4.3 \times 10^{-7} &= \frac{[\mathrm{HCO}_{3}^{-}](3.98\times 10^{-8})}{0.0012} \end{aligned}\] Now solve for \([\mathrm{HCO}_{3}^{-}]\): \[[\mathrm{HCO}_{3}^{-}] = \frac{4.3 \times 10^{-7} \times 0.0012}{3.98\times 10^{-8}} = 0.0013 M\]
04

Conclusion

The bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\) is \(0.0013 M\).

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.10 M\) acetic acid \(/ 0.25 M\) sodium acetate b. \(0.25 M\) acetic acid/0.10 \(M\) sodium acetate c. \(0.080 M\) acetic acid \(/ 0.20 M\) sodium acetate d. \(0.20 M\) acetic acid/0.080 \(M\) sodium acetate

Consider a buffered solution containing \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{CH}_{3} \mathrm{NH}_{2}\). Which of the following statements concerning this solution is(are) true? \(\left(K_{\mathrm{a}}\right.\) for \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}=2.3 \times 10^{-11}\).) a. A solution consisting of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) would have a higher buffering capacity than one containing \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\). b. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then the \(\mathrm{pH}\) is larger than the \(\mathrm{p} K_{\mathrm{a}}\) value. c. Adding more \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\right]\) to the initial buffer solution will decrease the \(\mathrm{pH}\). d. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]<\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}<3.36\). e. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}=10.64\).

A sample of a certain monoprotic weak acid was dissolved in water and titrated with \(0.125 \mathrm{M} \mathrm{NaOH}\), requiring \(16.00 \mathrm{~mL}\) to reach the equivalence point. During the titration, the \(\mathrm{pH}\) after adding \(2.00 \mathrm{~mL} \mathrm{NaOH}\) was \(6.912\). Calculate \(K_{\mathrm{a}}\) for the weak acid.

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 \(0.50 \mathrm{M} \mathrm{HCl}\), calculate the \(\mathrm{pH}\) under the following conditions. a. after \(50.0 \mathrm{~mL}\) of \(0.50 M \mathrm{HCl}\) has been added b. at the stoichiometric point

Could a buffered solution be made by mixing aqueous solutions of \(\mathrm{HCl}\) and \(\mathrm{NaOH}\) ? Explain. Why isn't a mixture of a strong acid and its conjugate base considered a buffered solution?

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