Complete the following table of Trial 1(See Report sheet) for determining the mo

Question

Complete the following table of Trial 1(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 experimental procedure. Record calculated values with the correct number of significant figures. 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 analyst, 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 D.

Explanation / Answer

1. Given the mass of KHC8H4O4 taken = 0.411 g

2.Molar mass of KHC8H4O4 = 204.44 g/mol

3.Hence moles of KHC8H4O4 = mass / molar mass = 0.411 g / 204.44g/mol

= 0.00201 mol

4. Initial buret reading = 4.20 mL

5. Final buret reading = 19.90 mL

6.Hence volume of NaOH dispensed = 19.90 mL – 4.20 mL = 15.70 mL

7. Let the molar concentration of NaOH be M

Hence moles of NaOH = MxV(L) = Mx15.70 mL x (1L / 1000mL) = 0.01570M mol

The neutralization reaction of KHC8H4O4 with NaOH is

of KHC8H4O4 + NaOH ---- > NaKC8H4O4 + H2O

Hence moles of NaOH = moles of of KHC8H4O4

=> 0.01570M mol = 0.00201 mol

=> M = 0.00201 / 0.01570 = 0.128 M

Molar concentration of NaOH solution = 0.128 M or 0.128 mol/L (answer)

B: Molar concentration of acid solution:

Either concentration or moles of NaOH required to solve this problem.

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