A buffer that contains 0.578 M base, B, and 0.263 M of its conjugate acid, BH , has a pH of 8.97. What is the pH after 0.0020 mol of HCl is added to 0.250 L of this solution?

pH = 8.95

Explanation:

A buffer follows H-H equation:

pH = pKa + log [B] / [BH⁺]

Where pH is pH of the buffer: 8.97

pKa is a constant of the buffer (Our incognite)

[B] is molar concentration -or moles- of the base: 0.578M

[BH⁺] is molar concentration -or moles- of conjugate acid: 0.263M

pH = pKa + log [B] / [BH⁺]

8.97 = pKa + log [0.578M] / [0.263M]

8.628 = pKa of the buffer

0.250L contains:

[B] = 0.578M * 0.250L = 0.1445 moles of B

[BH⁺] = 0.263M * 0.250L = 0.06575 moles of BH⁺

When HCl is added, B is consumed and BH⁺ is produced as follows:

HCl + B → BH⁺ + Cl⁻

That means moles of each species after the addition of HCl is:

[B] = 0.1445 moles - 0.0020moles = 0.1425moles

[BH⁺] = 0.06575 moles + 0.0020moles = 0.06775moles

Replacing in H-H equation, pH is:

pH = 8.628 + log [0.1425moles] / [0.06775moles]