Question

What is the change in pH when 10.0 mL of 0.250 M KOH (aq) is added to 100.0 mL of a buffer that is 0.150 M in HCOOH (aq) and 0.100 M in HCOO^– (aq)? pK_a = 3.75 for HCOOH (aq).

Select one:

a. –0.26 pH unit

b. +0.18 pH unit

c. +0.00 pH unit

d. –0.81 pH unit

QUESTION 2

What is the molarity of an aqueous barium hydroxide solution if it takes 59.9 mL of 0.100 M HCl (aq) to neutralize 21.6 mL of the Ba(OH)₂ (aq) solution?

Select one:

a. 0.208 M

b. 0.277 M

c. 0.139 M

d. 0.0693 M

Answer 1

What is the change in pH when 10.0 mL of 0.250 M KOH (aq) is added to 100.0 mL of a buffer that is 0.150 M in HCOOH (aq) and 0.100 M in HCOO^– (aq)? pK_a = 3.75 for HCOOH (aq).

Solution :-

Lets first calculate the moles of each

Moles of KOH = 0.250 mol per L * 0.010 L = 0.0025 mol

Moles of HCOOH = 0.150 mol per L * 0.100 L = 0.0150 mol

Moles of HCOO- = 0.100 mol per L * 0.100 L = 0.0100 mol

After adding the NaOH total volume = 100 ml + 100 ml + 10 ml = 210 ml = 0.210 L

NaOH reacts with HCOOH to produce the HCOO-

New moles of HCOOH = 0.0150 mol – 0.0025 mol = 0.0125 mol

New moles of HCOO- = 0.0100 mol + 0.0025 mol = 0.0125 mol

Moles of acid and base are same so the concnetrations are also same

Therefore the ratio of the concentration is 1

So the pH= pka

So the answer is

pH= 3.75

initial pH is calculated as

after mixing both solutions new concentration of the HCOOH and HCOO- are

[HCOOH] = 0.150 M * 100 ml / 200 ml = 0.075 M

[HCOO-] = 0.100 M * 100 ml / 200 ml = 0.050 M

pH = pka + log ([base]/[acid])

      = 3.75 + log [0.050 /0.075]

      = 3.57

So the change in the pH is

3.75-3.57 = 0.18

So the answer is option 'b'

QUESTION 2

What is the molarity of an aqueous barium hydroxide solution if it takes 59.9 mL of 0.100 M HCl (aq) to neutralize 21.6 mL of the Ba(OH)₂ (aq) solution?

Solution :-

Moles of HCl = 0.100 mol per L * 0.0599 L = 0.00599 mol

Mole ratio of the HCl to Ba(OH)2 is 2 : 1

So the moles of Ba(OH)2 = 0.00599 mol HCl * 1 mol Ba(OH)2/2mol HCl = 0.002995 mol

Molarity of the Ba(OH)2 =moles / volume in liter

                                         = 0.002995 mol / 0.0216 L

                                         = 0.139 M

So the answer is option c