NaOH Final reading - 16.8 mL 4. b. Complete the following table and show the cal

Question

NaOH Final reading - 16.8 mL

4. b. Complete the following table and show the calculations for the determinations for Trial (see Report Sheet) Record all calculated data to the correct number of significant figures. Calculation Zone 1. Mass of dry, solid acid (g) 0,297 Part 7 2. Molar concentration of NaOH (mol/L) -0.150 3. Buret reading of NaoH, initial (ml) 4. Buret ding of NaOH at stoichiometric point, final (mL) 5. Volume of N dispensed at stoichiometric point (ml, from graph) Part 8 7. Moles of NaOH to the point (mol) ometric Show calculation. 8. Moles of acid (mol Show calculation. Part 9 9. Molar mass of acid (g/mol Show calculation. 11. Volume of NaOH halfway to the stoichiometric point (ml, from graph) 12. pKa of weak acid (from graph)

Explanation / Answer

So,

4. Final NaOH = 16.8mL

5. NaOH from graph = 16.5mL ( at the equilvalence point)

7. Moles of NaOH to point = molarity x volume = 0.15M x 0.165L = 0.002475moles

8. Moles of acid = As NaOH and acid has 1:1 stoichiometry, moles of NaOH = moles of acid = 0.002475moles

9. molar mass of acid = mass/moles = 0.297/0.002475 = 120g/mol

11. Volume of NaOH halfway: 16.5/2 = 8.25mL

12 pKa = pH at 8.25mL of NaOH = approx 2.7

Have a nice day!!

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