Question

A 1.0 M solution of a compound with 2 ionizable groups (pKa's = 6.2 and 9.5; 100 mL total) has a pH of 6.8. If a biochemist adds 60 mL of 1.0 M HCl to this solution, the solution will change to pH

Answer 1

Answer:

pH = 9,32

Explanation:

The compound with 2 ionizable groups has the following equilibriums:

H₂M ⇄ HM⁻ + H⁺ pka = 6,2

HM⁻ ⇄ M²⁻ + H⁺ pka = 9,5

The reaction of M²⁻ with HCl is:

M²⁻ + HCl → HM⁻ + Cl⁻

The moles of M²⁻ are:

0,100L×1,0M = 0,1moles

And moles of HCl are:

0,060L×1,0M = 0,06moles

That means that moles of M²⁻ will be 0,1-0,06 = 0,04mol and moles of HM⁻ will be the same than HCl, 0,06mol

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [M²⁻] / [HM⁻]

Replacing:

pH = 9,5 + log₁₀ [0,04] / [0,06]

pH = 9,32

I hope it helps!