I am trying to complete a lab report, (k a of an indicator), but am stuck on how
ID: 527899 • Letter: I
Question
I am trying to complete a lab report, (ka of an indicator), but am stuck on how to find the value of NHAc /NAc could you explain how to fill out the table values below.(based on given info)
2 ml of 3.0 X 10^-4 M Bromcresol green soln
5ml of 1.6 M acetic acid (HAc)
2 ml of 0.200 M KCl
2 ml of 3.0 X10^-4 M Bromcresol green
2 ml of 0.160 M sodium acitate
here's the table
Please explain in detail
Soln A (diluted to 50 ml) Soln B (diluted to 50 ml)2 ml of 3.0 X 10^-4 M Bromcresol green soln
5ml of 1.6 M acetic acid (HAc)
2 ml of 0.200 M KCl
2 ml of 3.0 X10^-4 M Bromcresol green
2 ml of 0.160 M sodium acitate
Explanation / Answer
pH = -log [H+]
4.87 = -log [H+]
[H+] = Antilog -4.87 = 1.35 X 10-5 M
Consider dissociation of a weak acid
HC2H3O2 <-------> H+ + C2H3O2-
Concentrations at equilibrium can be calculated as follows:
[HC2H3O2]int = molarity X volume of acid / total volume of the solution
= 1.6 M X 5 mL / 50 mL
= 0.16 M
As the acetic acid starts dissociating, the concentration of [HC2H3O2] decreases and at equilibrium it is given as:
[HC2H3O2]e = [HC2H3O2]int - [H+]
= 0.16 - 1.35 X 10-5 = 0.1599865 M
Number of moles of HC2H3O2 = [HC2H3O2]e X Total volume of solution
When 1 mL of solution A is added to 50 mL of B
= 0.1599865 M X 51 X 10-3 L
= 8.1593 X 10-3 mol
[C2H3O2-]int = molarity X volume of Na acetate / total volume of the solution
= 0.16 M X 5 mL / 50 mL
= 0.016 M
At equilibrium the [C2H3O2-] will be from two sources namely Na acetate and dissociated acetic acid. Since the acetic acid is monoprotic acid [H+]: [C2H3O2-] = 1:1 and the generated [C2H3O2-] = [H+]
[C2H3O2-]e = [C2H3O2-]int + [H+]
= 0.016 + 1.35 X 10-5 = 0.0160135 M
Number of moles of C2H3O2-= [C2H3O2-]e X Volume of the solution B (You have not provided this volume)
= 0.0160135 M X 51 X 10-3 L
= 8.1667 X 10-4 mol
The ratio of Number of moles of HC2H3O2 : C2H3O2- = 8.1593 X 10-3 mol / 8.1667 X 10-4 mol
NHAc/ NAc- = 9.99
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