Question

REPOSTING: observing pH changes in Water and Buffer solutions.

the amount added of HCl and NaOH are both 25mL.

the buffer solution is made of 2grams of NaC2H3O2 and 4 mL of 6M HC2H3O2 in a 50 mL solution (46mL of water). the solution will contain 2.4x10^-2 mol each of NaC2H3O2 a d HC2H3O2.

question:

Lab 9: Studying the pH of Strong Acid, Weak Acid, Salt, and Buffer Solutions 1. Observing pH Changes in Water and Buffer Solutions upon Addition of HCl and NaOH Solutions theoretical ( HO) calculated pH solution measured pH 6.79 4.72 - 7.0 distilled water 10x/0-7 1.8x105 original buffer solution distilled water with added 6M HCI buffer solution with added 6M HCI 17:01 distilled water with added 6M NaOH buffer solution with added 6M NaOH 4.94 ngage Learning

how to answer it (I'm not sure I'm doing it right and want an expert to double check):

I need help calculating the theoretical and H3O+ concentration and the pH based off the theoretical.

Answer 1

1. For the distilled water, pH = -log[H+] = - log(1* 10-7) = 7

2. For your original buffer solution:

The pH of a buffer solution is calculated from the Henderson-Hasselbach equation which is given below.

salt pH = pka +log CH3COONa = pka + log! CH3O0H] Jacid

where, pH = -log[H+]

pKa = -log(Ka) = -log(1.78* 10-5) =4.75 ; Ka = acid dissociation constant of the weak acid (here CH3COOH)

[salt] = concentration of the salt (here CH3COONa)

  [acid] = concentration of the weak acid (here CH3COOH)

Now,

Molecular mass of CH3COONa = 82 g/mol

Therefore, 2g CH3COONa = (2/ 82) mol = 2.4* 10-2 mol CH3COONa.

2.4 x 10-2 mol CH3COONa] =? = 0.48 M

[CH3COOH] is calculated from the dilution principle, which is

V1 S1 = 125

where,

Vi = initial volume of acid = 4mL

  

Si = initial molarity of acid = 6 M

  

V2 = final volume of acid = 50 mL

  

S2 = final molarity of acid =?

Visi S2= 4 x 6 -M = 0.48 M _k 50

CH3COONa pH = pka +log" CH300H) -4.75 + log 10.48 0.481

pH = 4.75 + log1 = 4.75 -0 = 4.75 (answer)

Therefore,

SLT=l+H]601-

H+1 = 10-4.75 = 1.78 x 10-5 M (answer)

3. Distilled water with added 0.5 mL 6M HCl solution:

Since HCl is a strong acid, it will be completely dissociated and produce a solution of 6M H+ solution. Now, the contribution of water to the concentration of H+ is negligible compared to that produced by HCl.

We will again use the formula to calculate the final concentration of H+ in the solution.

S2 = Visi V2

V1 S 0.5 x 6, 1225.5 = M = 0.118 M

Therefore,

H+1=0.118 M

pH = -log[H+] = -log(0.118) = 0.93

4. For bufffer solution+0.5 mL 6M HCl solution:

0.5 mL 6M HCl contains of HCl

0.5 1000 x 6M = 3 x 10-3 mole

This will convert the 3* 10-3 moles of CH3COONa to form additional 3* 10-3 moles of CH3COOH.

Therefore, the concentrations after adding HCl are

(2.4 x 10-2 – 3 x 10-3) mol CH3COONa] =? -= 0.42 M

50 mL of 0.48M CH3COOH contains 2.4* 10-2 CH3COOH.

(2.4 x 10-2 +3 x 10-3) mol CH3COOH) = 4 = 0.54 M (500L)

Therefore,

CH3COONa pH = pka +log" CH300H) 5+log 10.42 9 10.54

pH = 4.75 +log(0.78) = 4.75 -0.11 = 4.64 (answer)

Ꭽ9Ꭽ =l+H]601-

H+1 = 10-4.64 = 2.29 x 10-5 M (answer)

5.Distilled water with added 0.5 mL 6M NaOH solution:

Since NaOH is a strong base, it will be completely dissociated and produce a solution of 6M OH- .

We will again use the formula to calculate the final concentration of OH- in the solution.

S2 = Visi V2

V1 S 0.5 x 6, 1225.5 = M = 0.118 M

Therefore,

OH-]=0.118 M

pOH = -log[OH-] = -log(0.118) = 0.93

pH = 14 – POH = 14 -0.93 = 13.07

pH = -log[H+] = 13.07

H+1= 10-13.07 = 8.51 x 10-14 M

6.For bufffer solution+0.5 mL 6M NaOH solution:

0.5 mL 6M NaOH contains of NaOH.

0.5 1000 x 6M = 3 x 10-3 mole

This will convert the 3* 10-3 moles of CH3COOH to form additional 3* 10-3 moles of CH3COONa.

Therefore, the concentrations after adding NaOH are

(2.4 x 10-2 +3 x 10) mol _ 0.54 M CH3COONa = ? (500L)

50 mL of 0.48M CH3COOH contains 2.4* 10-2 CH3COOH.

(2.4 x 10-2 - 3 10-3) mol CH3COOH) = 4 ' = 0.42 M 600 L)

Therefore,

CH3COON poh = pKa +log" CH300H = 4.75 +log 0.541 0.42]

pOH = 4.75 + log(1.286) = 4.75 +0.11 = 4.86

pH = 14 - pOH = 14 – 4.86 = 9.14

+16=l+H]601-

H+1= 10-9.14 = 7.25 x 10-10 M (answer)