Question

For each trial, compute the mol of titrant; (molarity x L) and keep the number of significant figures to 4.

Trial 1: 12.49 mL =
mol NaOH
Trial 2: 12.32 mL =
mol NaOH
Trial 3: 11.87 mL =
mol NaOH

Answer 1

Answers:

Trial 1: 12.49 mL = 0.0025 mol NaOH

Trial 2: 12.32 mL = 0.0025 mol NaOH

Trial 3: 11.87 mL = 0.0024 mol NaOH

Second step answers:

Trial 1 (20.61 mL HCl): 0.121 M

Trial 2 (20.06 mL HCl): 0.125 M

Trial 3 (19.67 mL HCl): 0.122 M

The average concentration (molarity) of HCl for all three trials is 0.123 M.

The percent error is 2.5%.

Answer 2

0.0025

0.0025

0.0024

Step-by-step explanation:

That is the correct answer