Question

Mixing Strong Acids with Strong Bases: Method for Solving for a ina 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 Keq values that are so large, we treat them as if they go to completion. In this case we can write the equation: HCl (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 HCl, 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 mixing the [HCl]-[H3O+] 0.25 moles 2 liters = 0.125 M. Remember the strong acid is "completely dissociated" in solution. We can then calculate the pH of the final solution using this [H3O+] 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 32.00 mL of 0.70 M HOI with 9.00 mL of 2.60 M NaOH? 1.61

Answer 1

Answer:

pH = 12.4

Explanation:

Following the hints that are given in this question, we can make a calculation based on the stoichiometry for the neutralization reaction:

HOI + NaOH    ⇒  NaOI + H₂O

We have to recognize that sodium hydroxide being such a strong base will react completely with the weak acid  HOI .

Lets determine the number of moles moles of each of the compounds involved in the reaction:

First we convert the volumes to liters for unit consistency.

Volume HOI = 32 mL x 1 L/1000 mL = 0.032 L

Volume NaOH = 9 mL x 1 L/1000 mL = 0.009 L

# moles HOI = 0.032 L x 0.70 mol HOI / 1L = 0.0224 mol

# moles NaOH = 0.009 L x 2.60 mol NaOH /1L = 0.0234 mol

We know from the stoichiometry of the reaction that 1 mol HOI will react with 1 mol NaOH, therefore 0.0224 mol HOI will react with 0.0224 mol NaOH and is the limiting reagent.

We have at the end of the mixing an excess of 0.001 mol NaOH which is the species that drives the pH being a strong base, even though we have the salt NaOI  which hydrolyzes in water, the NaOH is much more stronger and determines the pH.

Therefore we can ignore the effect on the pH of NaOI.

The problem now is the simple calculation of the pH of a strong base.

For the base we determine the pOH and then from the equation pH + pOH = 14 we will find the pH.

pH = - log [OH⁻]

mol OH⁻ = 0.001 mol

[OH⁻] = mol OH⁻/ Volume sol (L)

Vol solution = 0.032 L + 0.009 L  = 0.041 L

[OH⁻] = 0.001 mol / 0.041 L = 0.024 M

pOH = -log ( 0.024 ) = 1.6

pH = 14 - 1.6 = 12.4