Question

If you have 140. mL of a 0.150 M TRIS buffer at pH 8.30 and you add 3.00 mL of 1.00 M HCl, what will be the new pH? (The pK_a of TRIS is 8.30.)

Answer 1

solution:

We are given, molarity of TRIS = 0.15 M , volume = 0.140 M ,

pH = 8.30, volume of HCl = 3.00 mL , volume = 1.0 M

moles of HCl = 1.0 M * 0.003 L

                      = 0.003 moles

When we added the moles HCl there is moles of acid increase and moles of conjugate base decrease

We are given the pH = 8.30 and also given the pKa = 8.30

Means the [acid] = [conjugate base]

So, moles of acid = 0.15 M * 0.140 L = 0.021 moles

Moles of conjugate base = 0.15 M * 0.140 L = 0.021 moles

Moles of acid = 0.021 moles + 0.003 moles = 0.024

Moles of conjugate base = 0.021- 0.003 = 0.018 moles

So, assume volume 1.0 L

        pH = pKa + log [conjugate acid] / [acid]

so,

    = 8.30 + log 0.018/0.024

   = 8.17

   The new pH will be 8.17