A buffer is composed of 0.400 mol H₂PO₄^- and 0.400 mol HPO₄^2- diluted with water to a volume of 1.00 L.The pH of the buffer is 7.210.How many moles of HCl must be added to decrease the pH to 6.210? A) 0.200 mol B) 0.327 mol C) 0.360 mol D) 0.400 mol E) 3.60 mol

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Quizplus · Accepted Answer

To solve this problem, we will use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]) where pH is the current pH of the buffer (7.210), pKa is the acid dissociation constant of the acid in the buffer (we assume that HS- is acting as an acid and HPO42- is acting as its conjugate base), [A-] is the concentration of the conjugate base (HPO42-), and [HA] is the concentration of the acid (HS-). Rearranging the equation, we get: log([A-]/[HA]) = pH - pKa Substituting the values given in the problem, we get: log(0.400/0.400) = 7.210 - 6.950 log(1) = 0.260 Now, to decrease the pH to 6.210, we need to add HCl, which will react with the buffer and increase the concentration of H+ ions. Let's say we add x moles of HCl. This will react with the buffer as follows: HS- + H+ <--> H2S HPO42- + H+ <--> H2PO4- We can assume that x is small compared to the initial amount of buffer, so we can use the initial concentrations of HS- and HPO42- for the Henderson-Hasselbalch equation. We also assume that the volume of the solution doesn't change significantly upon addition of the HCl. The new pH will be: pH = pKa + log([A-]/[HA]) where the new [A-] and [HA] values are: [A-] = 0.400/(1 + 10^(pH-pKa)) [HA] = 0.400 - [A-] Substituting the given values for the initial buffer, we get: pKa = 6.950 [A-] = 0.200 mol/L [HA] = 0.200 mol/L Substituting the new pH of 6.210, we get: [A-] = 0.327 mol/L [HA] = 0.073 mol/L We can now solve for x: 6.210 = 6.950 + log(0.327/0.073) x = 0.360 mol Therefore, the answer is (C) 0.360 mol.

A Buffer Is Composed of 0