A solution of sodium hydroxide was standardized by titrating known volumes of th

Question

A solution of sodium hydroxide was standardized by titrating known volumes of the above KHP solution. The resulting data are summarized in the table below:

The concentration of the KHP solution is .06431.

Questions 2,3,5,6!

solution. The resulting data are summarized in the table below Write PDF file RUN 1 2.7 35.6 2.85 22.69 RUN 2 1.15 35.38 0.53 21.11 Open Print Friendly Window Initial KHP buret reading Final KHP buret reading Initial NaOH buret reading Final NaOH buret reading Change Quiz Choose a Problem Set What is the concentration (M) of the NaOH for Run 1 of the standardization? Question 2 Enter Your Answer: .11 Incorrect What is the concentration (M) of the NaOH for Run 2 of the standardization? Question 3 Enter Your Answer: 107 107 Correct, but too few significant figures What is the average concentration (M) of the NaOH for the above two runs? Question 4 [Note: The accuracy of the average concentration of NaOH is critical in the answers to most of the subsequent questions.] Enter Your Answer: 1085 Incorrect What is the average deviation of the concentration of the NaOH for the above two runs? Question 5 Enter Your Answer: What is the percent deviation of the concentration of the NaOH for the above two runs? Question 6

Explanation / Answer

# Balanced Reaction:            KHC8H4O4(aq) + NaOH(aq) -------> KNaC8H4O4(aq) + H2O(l)

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

Given, [KHP] = 0.06431 M

#2. Run 1: Volume of standard KHP solution consumed = 35.6 mL – 2.70 mL = 32.90 mL

Volume of NaOH solution consumed = 22.69 mL – 2.85 mL = 19.84 mL

Now,

            Using M1V1 (KHP) = M2V2 (NaOH)

            Or, 0.06431 M x 32.90 mL = M2 x 19.84 mL

            Or, M2 = (0.06431 M x 32.90 mL) / 19.84 mL = 0.10664 M

Hence, molarity of NaOH in Run 1 = 0.106643 M = 0.107 M

#3. Run 2: Volume of standard KHP solution consumed = 35.38 mL – 1.15 mL = 34.23 mL

Volume of NaOH solution consumed = 21.11 mL – 0.53 mL = 20.58 mL

Now,

            Using M1V1 (KHP) = M2V2 (NaOH)

            Or, 0.06431 M x 34.23 mL = M2 x 20.58 mL

            Or, M2 = (0.06431 M x 34.23 mL) / 20.58 mL = 0.106964 M

Hence, molarity of NaOH in Run 1 = 0.106964 M = 0.107 M

#4. Average molarity of NaOH = (Molarity in Run 1 + Molarity in Run 2) / 2

                                                = (0.106643 M + 0.106964 M) / 2

                                                = 0.1068035 M

#5. Average deviation = (I Molarity 1 – Avg. M I + I Molarity 2 – Avg. M I) /2

                                    = (I 0.106643 -0.1068035 I + I 0.106964 - 0.1068035 I) / 2

                                    = 0.0001605

Hence, average deviation in molarity = 0.0001605 M

Note: “ I Mode I “. Molarity 1, 2 = Molarity of NaOH in run 1 and 2, respectively.

#6. % Deviation = (Average deviation / Average molarity) x 100

                                    = (0.0001605 M / 0.1068035 M) x 100

                                    = 0.15 %

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