A. One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum b

Question

A.        One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that

One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that one solution is a buffer is not sufficient. of acetic acid = 1.8 x 10-5 Acetic acid/sodium acetate buffer Acetic acid .1M Sodium acetate .1M Convert 0.25 mL of 0.10M HCl->mols HCl Convert 10 mL of 0.1M HC2H3O2->mols HC2H3O2 Convert 10 mL of 0.1M C2H3O2->mols C2H3O2 Stoichiometry Calculations Convert 0.00975 mol C2H3O2 after reaction to concentration: Convert 0.001025 mol HC2H3O2 after reaction to concentration: Concentration Calculations: Solve for new pH: Alternatively, calculate pH using the following method: The addition of the acid changed the pH by 0.02 units Water: Add 0.25 mL of 0.1M HCl to 10 mL H2O Convert 0.25 mL of 0.10M HClmols HCl Concentration of H3O+ in 10.25 mL: Calculate new pH:

Explanation / Answer

1) moles of Hcl = molarity x volume /1000 = 0.10 x 0.25 /1000 = 2.5 x 10-5 moles

2) moles of CH3C00H = 10 x 0.1 /1000 = 10-3

3) moles of CH3C00- = 10 x 0.1 /1000 = 10-3

4) CH3C00- + H30+ -----------> CH3C00H + H20

moles of CH3C00H = 10-3 + 2.5 x 10-5 = 1.025 x 10-3

moles of CH3C00- = 10-3 - 2.5 x 10-5 = 9.75 x 10-4

5) total volume = 10 +0.25 =10.25 ml

conc of CH3COOH = moles x 1000/volume = 1.025 x 10-3 x 1000/10.25 = 0.1

6) COnc of CH3COO- = 9.75 x 10-4 x 1000/10.25 = 0.0951219 M

7) initial - 0.1 ,0. 0.1

chnage - x , x , - x

final = 0.1+ x , x , 0.1 - x

8) ka = x * ( 0.1+x) / (0.1 -x)

x = 1.8 x 10-5

[H30+] = 1.8 x 10-5 = Ka

9) pH = pKa + log 0.095/ 0.1

pH = 4.7447 -0.0227

pH = 4.722

10) moles of Hcl = .25 x0.1 /1000 = 2.5 x 10-5

11) conc of H30+ = 2.5 x 10-5 x 1000/10.25 = 2.439 x 10-3

12) pH = -log 2.439 x 10-3

pH = 2.612

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