Final answer:
Performing the calculations reveals that the final concentration of OH- ions after titrating 80 mL of 2 M NaOH with 2 mL of 4 M HCl is approximately 1.85 M. The options given in the question did not include the correct answer.
Step-by-step explanation:
Titration is a quantitative chemical analysis method used to determine the concentration of a known reactant in a solution. In the given scenario, the student is titrating 80 mL of 2 M NaOH with 2 mL of 4 M HCl. To find the final concentration of hydroxide ions (OH-), we need to calculate the amount of NaOH that reacts with HCl as well as the amount that remains unreacted.
First, we use the formula n = C × V where n is the number of moles, C is the concentration, and V is the volume in liters. For NaOH, nNaOH = 2 M × 0.080 L = 0.16 mol. For HCl, nHCl = 4 M × 0.002 L = 0.008 mol. Since NaOH and HCl react in a 1:1 molar ratio, all of the HCl will react with an equivalent amount of NaOH, leaving 0.16 mol - 0.008 mol = 0.152 mol of NaOH.
The total volume after the reaction is 80 mL + 2 mL = 82 mL or 0.082 L. The final concentration of OH- ions is therefore OHCfinal = 0.152 mol / 0.082 L ≈ 1.85 M.
The correct answer to what is the final concentration of OH- ions is not listed in the original options provided. It is 1.85 M.