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

The Henderson-Hasselbalch equation relates the pH, pKa, and the concentrations of a weak acid and its conjugate base: \[ pH = pK_{a} + \log{\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}} \] In this case, the acid is the carbonic acid, H2CO3, and the base is the bicarbonate ion, HCO3-.

We are given the Ka value, and we need to find the pKa: \[ pK_{a} = - \log{K_{a}} \] Substitute the given Ka: \[ pK_{a} = - \log{(4.3 \times 10^{-7})} \] Calculate the pKa value: \[ pK_{a} \approx 6.37 \]

We are given the pH value, which is 7.40, and we have calculated the pKa value to be approximately 6.37. We are also given the carbonic acid concentration, which is 0.0012 M. We can substitute these values into the equation: \[ 7.40 = 6.37 + \log{\frac{[\mathrm{HCO}_{3}^{-}]}{[0.0012]}} \]

We want to find the value of [HCO3-], so we need to isolate it in the equation. First, subtract the pKa value from both sides: \[ 1.03 = \log{\frac{[\mathrm{HCO}_{3}^{-}]}{[0.0012]}} \] Now we need to find the antilogarithm of both sides to get rid of the logarithm: \[ 10^{1.03} = \frac{[\mathrm{HCO}_{3}^{-}]}{[0.0012]} \] Calculate the antilogarithm: \[ \approx 10.68 = \frac{[\mathrm{HCO}_{3}^{-}]}{[0.0012]} \] Now, multiply both sides by the carbonic acid concentration: \[ [0.0012] \times 10.68 = [\mathrm{HCO}_{3}^{-}] \] Calculate the bicarbonate ion concentration: \[ [\mathrm{HCO}_{3}^{-}] \approx 0.0128 \,M \] So, to buffer the pH of blood at 7.40, the bicarbonate ion concentration needs to be approximately 0.0128 M.