Question

Use the results of the calculations belowe and literature values of K_a to predict the pH of each of the three buffer solutions before they are titrated.

Buffer 1

Moles of acid = 0.50 M x 0.050L = 0.025 moles

Moles of NaOH = 0.25M x 0.050L = 0.0125 moles

Final vol. of solution = 50 mL acid + 50 mL NaOH = 100 mL = 0.1 L

Initial concentration of weak acid = 0.025 moles/0.1L = 0.250M

Initial concentration of NaOH = 0.0125 moles/0.1L = 0.125M

                  Buffer 2

Moles of acid = 0.50 M x 0.050L = 0.025 moles

Moles of NaOH = 0.25M x 0.035L = 0.00875 moles

Final vol. of solution = 50 mL acid + 35 mL NaOH + 15 mL H₂O = 100 mL = 0.1 L

Initial concentration of weak acid = 0.025 moles/0.1L = 0.250M

Initial concentration of NaOH = 0.0875 moles/0.1L = 0.0875M

                  Buffer 3

Moles of acid = 0.50 M x 0.050L = 0.025 moles

Moles of NaOH = 0.25M x 0.025L = 0.00635 moles

Final vol. of solution = 50 mL acid + 25 mL NaOH + 25 mL H₂O = 100 mL = 0.1 L

Initial concentration of weak acid = 0.025 moles/0.1L = 0.250M

Initial concentration of NaOH = 0.00625 moles/0.1L = 0.0625M

Answer 1

Buffer 1

concentration of Acid =0.250 M

concentration of NaOH = 0.125 M

Here acid is excess, the amount of excess acid = 0.250-0.125 = 0.125 M

now equal molar acid reacts with equal molar NaOH, then [salt] = 0.125 M

using Henderson equation for acid buffer

pH = Pka + log [salt]/ [acid]

pH = pKa

Buffer 2

Intial [acid] =0.250M

Initial [NaOH] = 0.0875 M

after neutralisation reaction [acid] = 0.250-0.0875 =0.1625 M

[salt] = 0.0875 M

using Henderson equation for acid buffer

pH = Pka + log [salt]/ [acid]

pH = pKa+ log 0.0875/0.1625 = pKa + log 0.538

or, pH = pKa -0.268

Buffer 3

Intial [acid] =0.250M

Initial [NaOH] = 0.0625 M

after neutralisation reaction [acid] = 0.250-0.0625 =0.1875 M

[salt] = 0.0625 M

using Henderson equation for acid buffer

pH = Pka + log [salt]/ [acid]

pH = pKa+ log 0.0625/0.1875 = pKa + log 0.333

or, pH = pKa -0.477