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!
The solution will have a pH that is mathematically given as
pH = 9.32
Question Parameters:
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
Generally,The compound with 2 ionizable groups are
H_2M ⇄ HM⁻ + H⁺
with a pka of 6.2
HM⁻ ⇄ M²⁻ + H⁺
with a pka of 9.5
The chemical reaction is
M^{2--} + HCl --> HM- + Cl-
Therefore, using Henderson-Hasselbalch formula whic is
pH = pka + log₁₀ [M²⁻] / [HM⁻]
pH = 9.5 + log₁₀ [0.04] / [0.06]
pH = 9.32
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