6. a. Complete the following table of Trial I (See Report Sheet) for determining

Question

6. a. Complete the following table of Trial I (See Report Sheet) for determining the molar concentration of a standard NaOH solution, followed by the determination of the molar concentration of a monoprotic acid solution according to the exper- imental procedure. Record calculated values with the correct number of significant figures A. Standardization of a Sodium Hydroxide Solution Calculation Zone 1. Tared mass of KHC,H,0, (8) 2. Molar mass of KHC,H404 (g/mol) 3. Moles of KHC,H,O, (mol) 0444 Show calculation. 4. Buret reading, initial (mL) 5. Buret reading. final (mL) 6. Volume of NaOH dispensed (ml.) 7. Molar concentration of Part A.7 NaOH solution (molVL) Show calculation B. Molar Concentration of an Acid Solution Part B.6 250- 3.20- 20.47 1. Volume of acid solution (mL.) 2. Buret reading, initial (mL) 3. Buret reading, final (mL) 4. Volume of NaOH dispensed (mL) 5. Molar concentration of NaOH solution (moUL)Part B.7 6. Moles of NaOH dispensed (mol) Show calculation. 7. Molar concentration of acid solution (moV/L) Show calculation. 6. b. For Trials 2 and 3, the molar concentration of the acid was 0.922 M and 0.856 M respectively a. What is the average molar concentration of the acid solution? Data Analysis, B. b What are the standard deviation and the relative standard deviation (%RSD) for the molar concentration of the acid solution? Data Analysis, C and ID

Explanation / Answer

A. Standardization of a NaOH solution:

1. Tared mass of KHC8H4O4 = 0.411g

2. MOlar mass of KHC8H4O4 = 204.44 g/mol

3. Moles of KHC8H4O4 = Mass of KHC8H4O4 / Molar mass of KHC8H4O4

= 0.411 g / 204.44g/mol = 2.01 x 10^-3 mol

4. Buret reading, initial = 4.20 mL

5. Buret reading, final = 19.90 mL

6. Volume of NaOH dispensed = Final - Initial = 19.90 - 4.20 = 15.7 mL

7. Molar concentration of NaOH solution (mol /L)

KHC8H4O4 (aq) + NaOH (aq) ? NaKC8H4O4 (aq) + H2O

1 mol of KHC8H4O4 reacts with 1 mol of NaOH

2.01 x 10^-3 mol of KHC8H4O4 reacts with "x" mol of NaOH

x = 2.01 x 10^-3 mol

Molarity = Moles / Volume(in L) = 2.01 x 10^-3 mol / (0.0157 L) = 0.128 M

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B. Molar concentration of Acid solution

1. Volume of acid solution = 25.0 mL

2. Buret reading, initial = 3.70 mL

3. Buret reading, final = 20.47 mL

4. Volume of NaOH dispensed = final - initial = 20.47 - 3.70 = 16.77 mL

5. Molar concentration of NaOH solution = 0.128 M

6. Moles of NaOH dispensed

Moles = Molarity x Volume (in L ) = 0.128 M x 0.01677 L = 2.14 x 10^-3 mol

7. Molar concentration of acid solution

Since, the given acid is a monoprotic acid, it reacts in equimolar ratio with NaOH.

Moles of acid reacted = 2.14 x 10^-3 mol

Molarity = Moles of acid / Volume of acid (in L) = (2.14 x 10^-3 mol) / 0.025 L = 0.0858 M

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6. b.

a. Average molar concentration of acid

(0.0858 + 0.922 + 0.856) / 3 = 0.621 M

b.

Standard deviation = square root (1/N[Sigma(xi - mean)^2]

Sigma(xi - mean)^2 = (0.0858-0.621)^2 + (0.922 - 0.621)^2 + (0.856 - 0.621)^2 = 0.432

square root (1/N[Sigma(xi - mean)^2] = square root(0.432 /3) = 0.379

Standard deviation = 0.379

Relative standard deviation = standard deviation / mean = 0.379 / 0.621 = 0.611

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