Solutions used: 0.100 M NaOH, 0.100 M HCl, 0.100 M HC 2 H 3 O 2 For Data sheet 1

Question

Solutions used: 0.100 M NaOH, 0.100 M HCl, 0.100 M HC2H3O2

For Data sheet 1& 2 find the pH, [H3O+], and [OH-]:

1) before adding NaOH solution

2) at equivalence point

3) after last addition

**Mainly need help figuring out the pH for the equivalence point.**

DATA SHEET 1 Titrating HCl solution with NaoH solution Buret Volume NaOH pH reading, mL added, mL 0 mL 2.11 0.00 1 mL 2.17 00 2 mL 2.27 2.00 3 mL 2.13 3.00 4 mL 2.67 4.00 5 mL 3.27 5.0d 6 mL, 23 6.00 7 mL 7.00 8 mL 55 8.00 9 mL 95 9.00 10 mL 0.00 12.08 11 mL li.00 2.IS 12 mL 12.00 12.22 13 mL IS oo 12.11 14 mL 12.32 4.00 15 mL 12.36 I5.00 L Titrating HCI solution with NaoH solution IH30 10H1 pH (1) before adding NaoH solution (2) at equivalence point (3) after last addition

Explanation / Answer

1)

before adding NaoH solution

Here are the equations you could use.

pH + pOH=14

pH =-log[H3O+]   [H3O+]=10-pH

pOH=-log[OH-]    [OH-]=10-pOH

Kw=1.0 x 10-14 = [H3O+] [OH¯]

If pH= 2.11 and pOH=14-2.11 pOH=14-2.11=11.89

    [H3O+]=10-pH   so [H+]=10-2.11    [H3O+]=7.7x10-3M

    [OH-]=10-pOH so [OH-]=10-11.89 [OH-]=1.28x10-12M

At equivalent point PH is 7.0

so pH = pOH=7

[H+] = [OH-]= 10-7 = 1.0 * 10-7

At end point PH is 12.36

pOH=14-12.36=1.64

[H+]=10-pH   so [H+]=10-12.32

[H+]=4.36x10-13M

[OH-]=10-pOH so [OH-]=10-1.64 [OH-]=2.2x10-2M

For Second Data sheet

2)

before adding NaoH solution

Here are the equations you could use.

pH + pOH=14

pH =-log[H3O+]   [H3O+]=10-pH

pOH=-log[OH-]    [OH-]=10-pOH

Kw=1.0 x 10-14 = [H3O+] [OH¯]

If pH= 3.48 and pOH=14-3.48 pOH=14-2.11=10.52

    [H3O+]=10-pH   so [H+]=10-3.48    [H3O+]=3.3x10-4M

    [OH-]=10-pOH so [OH-]=10-10.52 [OH-]=3.01x10-11M

At equivalent point PH is 7.0

so pH = pOH=7

[H+] = [OH-]= 10-7 = 1.0 * 10-7

At end point PH is 12.32

pOH=14-12.32=1.68

[H+]=10-pH   so [H+]=10-12.32

[H+]=4.78x10-13M

[OH-]=10-pOH so [OH-]=10-1.68 [OH-]=2.08x10-2M

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