Mixing Strong Acids with Strong Bases: Method for Solving for a Final pH When a
ID: 545779 • Letter: M
Question
Mixing Strong Acids with Strong Bases: Method for Solving for a Final pH When a solution of a strong acid and a strong base are mixed, they will neutralize each other, usually to form water and a salt. If there is an excess of either reagent, the pH of the final mixed solution will be determined by the concentration of this excess reagent in the final mixed solution. Thus, it is really an example of a limiting reagent problem. We must first figure out which is the limiting reagent. Then we can decide what is left over and what is its concentration after mixing (a dilution will occur during mixing). This concentration will allow us to quickly calculate the final pH. Imagine that 3 moles of HCI and 2.75 moles of NaOH were mixed to a total volume of 2 liters. Strong acids and strong bases react completely. While all reactions can be considered equilibria, strong acid/base reactions have Keg values that are so large, we treat them as if they go to completion. In this case we can write the equation: HCI (aq) + NaOH (aq) NaCl (aq) and H2O In our example, the limiting reagent is NaOH (it can produce fewer moles of product than the HCI). Each mole of NaOH, neutralizes one mole of HCI, so at the end of the reaction all of the NaOH is gone and there is 0.25 moles of unreacted HCI remaining. This means that after mising the (HC) -H0)-2moe the final solution using this [HO 0.25 moles .125 M. Remember the 0.125 M. Remember the strong acid is "completely dissociated" in solution, We can then calculate the pH of 2 liters We are almost never told the number of moles of each reagent directly, often we are given the volume of a solution of known concentration. If we remember that Molarity times volume - moles of solute, we can quickly determine the moles of each reactant and follow the same logic as before. Remember to divide by the total volume of the combined solutions because this will act to dilute the excess acid or base during the reaction. What is the ph of the solution that results from mixing 24.00 mL of 1.00 M HC) with 10.00 mL of 2.60 M NaOH?Explanation / Answer
Number of moles of HCl = 24 * 10^-3 L* 1 mol / L = 0.024 moles
number of moles of NaOH = 10 * 10^-3 L * 2.6 mol / L = 0.026 moles
HCL + NaOH ------> NaCl + H2O
1 mole of HCl requires 1 mole of NaOH for the completion of the reaction.
0.024 moles of HCl requires 0.024 moles of NaOH
finally remain moles of acid or base = 0.026 - 0.024 = 0.002 moles of NaOH
pOH of NaOH = -log[OH-] ( [OH-] = number of moles of OH-/total volume of the solution)
pOH = -log(0.002/(0.024+0.001))
pOH = 1.097
pH = 14 - pOH
pH = 14 - 1.097 = 12.003