Question

Your classmate finds a bottle simply marked HCl in the laboratory and decides to conduct an acid-base titration using a pH meter. They add 0.2 N sodium hydroxide (NaOH) to 25 mL of the HCl solution. The next day another lab-user tells you that they prepared the HCl solution in question and that it has a molarity of 0.15 M. What may explain the discrepancy between the experimental and theoretical values?

From the curve, pH at equivalence point is 7 and Volume of NaOH= 22ml

Answer 1

Ans. #Step 1: Experimental value: 1 mol HCl is neutralized by 1 mol NaOH.

Given- strength of NaOH = 0.2 N = 0.2 M

At titration endpoint, the moles of HCl must be equal to that of NaOH.

So,

            M1V1 (HCl) = M2V2 (NaOH)

            Or, M1 = (0.20 M x 22.0 mL) / 25.0 mL = 0.176 M

Therefore, experimental molarity of HCl = 0.176 M

# The most plausible discrepancy is-

The most common error might be associated with the strength of NaOH solution. NaOH solution is hydroscopic, so is NOT used as a primary standard. NaOH must be freshly prepared, then standardized with a primary standard (like KHP), and then only can be used to determine the molarity of test HCl solution.

# If the NaOH solution were old (though standardized against KHP), its molarity would decrease due to absorption of moisture (hygroscopic nature of NaOH solution). So, a lower molarity of NaOH would give a higher molarity of HCl (0.176 M instead of 0.15 M).

# If 0.20 N NaOH were freshly prepared from pellets (hygroscopic), but not standardized against the primary standard, the resultant solution would have a lower molarity than the expected molarity of 0.2 M because the pellets were contaminated with moisture. Again, lowering NaOH molarity would yield a higher than actual molarity of HCl solution in question.

Note: There may be many more errors arising at each and every step (from solution preparation to titration), but the most common issue is presented.