Question

Each symbol in the box below represents a mole of a component in one liter of a buffer solution; \(\bigcirc\) represents the anion \(\left(\mathrm{X}^{-}\right), \square 0=\) the weak acid \((\mathrm{HX}), \square=\mathrm{H}^{+},\) and \(\Delta=\mathrm{OH}^{-}\). Water molecules and the few \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions from the dissociation of \(\mathrm{HX}\) and \(\mathrm{X}^{-}\) are not shown. The box contains \(10 \mathrm{~mol}\) of a weak acid, \(\square 0,\) in a liter of solution. Show what happens upon (a) the addition of \(2 \mathrm{~mol}\) of \(\mathrm{OH}^{-}(2 \Delta)\). (b) the addition of \(5 \mathrm{~mol}\) of \(\mathrm{OH}^{-}(5 \Delta)\). (c) the addition of \(10 \mathrm{~mol}\) of \(\mathrm{OH}^{-}(10 \Delta)\). (d) the addition of \(12 \mathrm{~mol}\) of \(\mathrm{OH}^{-}(12 \Delta)\). Which addition (a)-(d) represents neutralization halfway to the equivalence point?

Short answer

Answer: Case B (Addition of 5 moles of OH-) represents neutralization halfway to the equivalence point.

Step-by-step solution

Case A: Addition of 2 moles of OH-

Initially, we have 10 moles of the weak acid (HX) and 0 moles of X-. When 2 moles of OH- are added, they will react with HX to form X- and H2O. The reaction can be represented as follows: HX + OH- → X- + H2O Since 2 moles of OH- are added, we can assume that 2 moles of HX will be consumed, and 2 moles of X- will be formed. Therefore, the new concentrations are: HX: 10 - 2 = 8 moles X-: 0 + 2 = 2 moles OH-: 0

Case B: Addition of 5 moles of OH-

For the addition of 5 moles of OH-, the reaction will proceed as follows: HX + OH- → X- + H2O Since 5 moles of OH- are added, we can assume that 5 moles of HX will be consumed, and 5 moles of X- will be formed. Therefore, the new concentrations are: HX: 10 - 5 = 5 moles X-: 0 + 5 = 5 moles OH-: 0

Case C: Addition of 10 moles of OH-

For the addition of 10 moles of OH-, the reaction will proceed as follows: HX + OH- → X- + H2O Since 10 moles of OH- are added, we can assume that 10 moles of HX will be consumed, and 10 moles of X- will be formed. Therefore, the new concentrations are: HX: 10 - 10 = 0 moles X-: 0 + 10 = 10 moles OH-: 0

Case D: Addition of 12 moles of OH-

For the addition of 12 moles of OH-, the reaction will proceed as follows: HX + OH- → X- + H2O Since 12 moles of OH- are added, all 10 moles of HX will be consumed, and 10 moles of X- will be formed. The remaining 2 moles of OH- will stay in the solution. Therefore, the new concentrations are: HX: 10 - 10 = 0 moles X-: 0 + 10 = 10 moles OH-: 2 moles

Halfway to the Equivalence Point

The equivalence point is the point at which the moles of the added base (OH-) are equal to the moles of the weak acid (HX). In this case, the equivalence point occurs when 10 moles of OH- are added. Halfway to the equivalence point would be when 5 moles of OH- are added. So, the addition that represents neutralization halfway to the equivalence point is case B.