HCl + NaOH yields NaCl + H2O In determining the concentration of an unknown HCl

Question

HCl + NaOH yields NaCl + H2O In determining the concentration of an unknown HCl solution we perform a titration of the unknown (plus indicator) with a solution of NaOH of known concentration. When we reach neutralization (as indicated by the indicator) all of the acid is used up and we can determine the concentration of the unknown acid. Answer the following questions: (a) Using 1M NaOH what is the concentration of the unknown HCl if we needed 50 ML of NaOH to neutralize 100 ML of unknown. (b) Would the calculated molarity of the unknown acid solution be higher, lower, or not affected if each of the following occurred. Explain your answers. I. The buret containing NaOH was rinsed with distilled water but not rinsed with NaOH before being filled. II. Your lab instructor miscalculated the Molarity of the NaOH solution you used in the titration (the true molarity was higher than calculated by your instructor) III. The tip of the NaOH buret contained an air bubble at the beginning of the titration but not at the end. (c) Suppose you used an unknown H2SO4 concentration instead of HCl. What change world need to be made to the calculation of the concentration of the acid.

Explanation / Answer

Ans. Balanced reaction:        HCl(aq) + NaOH(aq) ----> NaCl(aq) + H2O(aq)

Stoichiometry: 1 mol NaOH completely reacts with 1 mol HCl.

At equivalence point-

            M1V1 (of NaOH) = M2V2 (of HCl)        - equation 1

                        Where, M and V represent molarity and volume of respective solutions.

Putting the values in above equation-

            1.0 M x 50.0 mL = M2 x 100.0 mL

            Or, M2 = (1.0 M x 50.0 mL) / 100.0 mL = 0.5 M

Thus, molarity of unknown HCl solution = 0.5 M

Ans. #B1. The burette is rinsed with the solution that is to be filled in for titration. Rinsing the burette with NaOH removes any impurities (like, water or traces of any other solution that was previously filled in the burette) that could have been left behind in it during previous use.

If nor rinsed with NaOH after rinsing with water, presence of water at the inner wall of burette will dilute NaOH. Thus, the volume of NaOH required to get the endpoint will be increased due to lower concentration of NaOH solution in burette.

Increase in NaOH volume required (V1 becomes > 50.0 mL) will further increase the value of molarity of HCl (M2).

Note that M1 is directly proportional to M2 (equation 1). So, M2 increases with M1 and vice-versa

#B2. Making calculation with lower value of NaOH molarity (M1 is lowered) also decrease the concertation of HCl (M2).

#B3. Presence of air bubble at the tip of burette interferes with the final reading. The final burette reading would be greater than the actual value.

            Final reading (erroneous) = Actual final reading + Volume of Air bubble at tip.

So, volume of NaOH required to reach endpoint increases.

Therefore, it gives increased value of HCl concentration [see #B1]

#B4. H2SO4(aq) + 2NaOH(aq) ----> Na2SO4Cl(aq) + 2H2O(aq)

Stoichiometry: 1 mol NaOH completely reacts with 2 mol H2SO4.

At equivalence point-

            M1V1 (of NaOH) = 2 x M2V2 (of H2SO4)         - equation 2

M2 (for H2SO4) will be half of M2 (for HCl) for the same titration.

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