Acid base titration - # 2 calculate the molarity of the sodium hydroxide solutio

Question

Acid base titration -
# 2 calculate the molarity of the sodium hydroxide solution for each run ? Acid base titration -
# 2 calculate the molarity of the sodium hydroxide solution for each run ?
#4 calculate the molar mass of the unknown solid acid from each run.

14-8 Name. Saal Sartre Experiment 14 Acid-Base Titrations Data Part I. Standardization of the Sodium Hydroxide Solution Drawer Number Mass of weighing bottle + sample Mass of weighing bottle - sample Mass of HoC2O4 2 H2O Volume of HaC:O, solution 2.3 1349 250.00 mL Run Number Volume of oxalic acid used 25.00 mL 25.00 mL 25.00 L25.00 mL NaOH buret: final reading TONL NaOH buret: initial reading M | 3"-391mL 37.3~! Volume of NaOH used Part II. Determination of the Molar Mass of an Acid Unknown # n- Run Number Mass of test tube + sample Mass of test tube - sample Mass of sample,1390 NaOH buret NaOH buret 23.3L 23.ss 21.80 ML Initial reading O.Lon Volume of NaOH used

Explanation / Answer

Ans. Part I- Standardization of NaOH-

Moles of H2C2O4.2H2O = Mass / Molar mass

= 2.3239 g / (126.06604 g/ mol) = 0.018434 mol

1 Mol H2C2O4.2H2O is equivalent to 1 mol H2C2O4.

Now,

Molarity of H2C2O4 = Moles of H2C2O4 / Volume of solution in liters

                                    = 0.018434 mol / 0.250 L

                                    = 0.073736 M

# Balanced Reaction:            2 NaOH(aq) + H2C2O4(aq) ----> Na2C2O4 + 2 H2O

According to the stoichiometry of balanced reaction, 1 mol H2C2O4 neutralizes 2 mol NaOH.

Trial 1: Using             C1V1 (NaOH) = 2 x C2V1 (H2C2O4)           1.8434

            Or, C1 = 2 x (0.073736 M x 25.0 mL) / 37.0 mL = 0.09964 M

Therefore, [NaOH] in trial 1 = 0.00964 M

Trial 2: C1 (NaOH) = 2 x (0.073736 M x 25.0 mL) / 37.2 mL = 0.09911 M

Trial 3: C1 (NaOH) = 2 x (0.073736 M x 25.0 mL) / 37.3 mL = 0.09884 M

# Average NaOH molarity = (0.09964 + 0.09911 + 0.09884) M / 3

                                                = 0.0992 M

# Part II: Estimation of Molar mass of acid:

It’s assumed that the value “n=2” indicates that the acid is DIPROTIC.

Balanced reaction: H2A + NaOH ------------> Na2A + H2O

According to the stoichiometry of balanced reaction, 1 mol H2C2O4 neutralizes 2 mol NaOH.

# Trial 1: Moles of NaOH consumed = Average molarity x Volume consumed in liters

                                                            = 0.0992 M x [ (23.3 – 0.20) / 1000] L

                                                            = 0.00229152 mol

So, moles of acid in sample = (1/ 2) x Moles of NaOH consumed

                                                = (1/ 2) x 0.00229152 mol

                                                = 0.00114576 mol

# Now,

            Molar mass of acid = Mass of acid sample / Moles of acid

                                                = 0.1340 g / 0.00114576 mol

                                                = 116.953 g/mol

# Trial 2: Moles of NaOH = 0.0992 M x [ (23.55 – 0.00) / 1000] L = 0.00233616 mol

So, moles of acid in sample =(1/ 2) x 0.00233616 mol = 0.00116808 mol

# Molar mass of acid =0.1349 g / 0.00116808 mol = 115.489 g/mol

# Trial 3: Moles of NaOH = 0.0992 M x [ (22.80 – 0.00) / 1000] L = 0.00226176 mol

So, moles of acid in sample =(1/ 2) x 0.00226176 mol = 0.00113088 mol

# Molar mass of acid =0.1326 g / 0.00113088 = 117.254 g/mol

# Average molar mass of acid = (116.953 + 115.489 + 117.254) g mol-1 / 3

                                                = 116.565 g / mol

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