Calculate the pH of each of the solutions and the change in pH to 0.01 pH units

Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.50-M HCl to 450. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing alpha the change is negative. water pH before mixing = pH after mixing = pH change = 0.129 M C_2H_3O_2 1- pH before mixing = pH after mixing = pH change = 0.128 M HC_2H_3O_2 pH before mixing = pH after mixing = pH change = a buffer solution that is 0.129 M in each C_2H_3O_2^1- and HC_2H_3O_2 pH before mixing = pH after mixing = pH change =

Explanation / Answer

(b) 0.129 M CH3COO-

CH3COO- + H2O = CH3COOH + OH-

Ka of CH3COOH = 1.8 x 10-5 =

Kb = Kw/Ka

= (1.0 x 10-14) / (1.8 x 10-5)

= 5.56 x 10-10

Kb = [OH-] [CH3COOH] / [CH3COO-]

or 5.56 x 10-10 = (x) (x) / (0.129 - x)

or 5.56 x 10-10 = (x) (x) / (0.129) since Ka is very small, x term in the denominator can be neglected

or x2 = 7.17 x 10-11

or x = 8.47 x 10-6

So, [OH-] = x = 8.47 x 10-6 M

pOH = - log[OH-] = - log (8.47 x 10-6) = 5.07

pH = 14 - pOH = 14 - 5.07 = 8.93

Now,

pH = 8.93

or - log[H+] = 8.93

or [H+] = 10-8.93 = 1.17 x 10-9 M

10 mL of 3.5 M HCl

Moles of HCl = 0.01 L x 3.5 M = 0.035 moles

Total volume = 10 mL + 450 mL = 460 mL = 0.46 L

[HCl] = 0.035 moles / 0.46 L = 0.076 M

So, [H+] =  0.076 M

Total [H+] =  0.076 M + 1.17 x 10-9 M = 0.076 M

pH = -log  [H+] =  - log (0.076) = 1.12

Change in pH = 1.12 - 8.93 = -7.81

(c) 0.129 M CH3COOH

CH3COOH = H+ + CH3COO-

Ka = [H+] [CH3COO-] / [CH3COOH]

or 1.8 x 10-5 = (x) (x) / (0.129 - x)

or 1.8 x 10-5 = (x) (x) / (0.129) since Ka is very small, x term in the denominator can be neglected

or x2 = 2.3 x 10-6

or x = 1.52 x 10-3

So, [H+] = x = 1.52 x 10-3 M

pH = - log [H+] = - log (1.52 x 10-3) = 2.82

10 mL of 3.5 M HCl

Moles of HCl = 0.01 L x 3.5 M = 0.035 moles

Total volume = 10 mL + 450 mL = 460 mL = 0.46 L

[HCl] = 0.035 moles / 0.46 L = 0.076 M

So, [H+] =  0.076 M

Total [H+] =  0.076 M + 0.00152 M = 0.07752 M

pH = -log  [H+] =  - log (0.07752) = 1.11

Change in pH = 1.11 - 2.82 = -1.71

(d) Using Henderson-Hesselbalach equation

pH = pKa + log { [salt] / [acid]

= pKa + log { [CH3COO-] / [CH3COOH]

= 4.75 + log (0.129/0.129)

= 4.75 + 0

= 4.75

Moles of CH3COO- = 0.129 M x 0.45 L = 0.05805 moles

Moles of CH3COOH = 0.129 M x 0.45 L = 0.05805 moles

Now, moles of HCl = 3.5 M x 0.01 L = 0.035 moles

After adding 0.035 moles of HCl, it will react with 0.035 moles of CH3COO- to produce 0.035 moles of CH3COOH . Hence moles of CH3COO- will decrease and moles of CH3COOH will increase.

Moles of addition of HCl are

Moles of CH3COO- = 0.05805 moles - 0.035 moles = 0.02305 mole

Moles of CH3COOH = 0.05805 moles + 0.035 moles = 0.09305 mole

Total volume = 450 mL + 10 mL = 460 mL = 0.46 L

So,

[CH3COO-] = 0.02305 mole / 0.46 L = 0.05

[CH3COOH] = 0.09305 mole / 0.46 L = 0.20

So,

pH = pKa + log { [CH3COO-] / [CH3COOH]

= 4.75 + log (0.05/0.20)

= 4.75 + log (0.25)

= 4.75 - 0.6

= 4.15

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.