Question

40.0 ml of 1.45 m hydroiodic acid was added to 60 ml of 0.80 sodium hydroxide. identify the limiting reactant. determine the ph of the final solution. what must be done to the final solution to bring the final ph to 7?

Answer 1

NaOH is the limiting reactant in the reaction between HI and NaOH. The pH of the final solution is 1.00. To reach a pH of 7, we must add a base to neutralize the excess acid.

Step-by-step explanation:

To identify the limiting reactant, we must compare the mole ratio of hydroiodic acid (HI) and sodium hydroxide (NaOH) using the provided concentrations and volumes.

First, we calculate the moles of each reactant.

For HI: 1.45 M x 0.040 L = 0.058 mol.

For NaOH: 0.80 M x 0.060 L = 0.048 mol.

The reaction between HI and NaOH is a 1:1 stoichiometry:

HI + NaOH → NaI + H2O

Since we have more moles of HI (0.058 mol) compared to NaOH (0.048 mol), NaOH is the limiting reactant.

Next, we determine the pH of the final solution by calculating the excess moles of HI (0.058 mol - 0.048 mol = 0.010 mol).

The concentration of excess HI is 0.010 mol / (0.040 L + 0.060 L) = 0.100 M.

Since HI is a strong acid, the pH is equal to -log[0.100] = 1.00.

To bring the final pH to 7, we must neutralize the excess acid by adding a strong base, such as NaOH, until the solution is neutral.