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What volume, in milliliters, of 0.100 M NaOH should be added to a 0.135 L solution of 0.017 M glycine hydrochloride (p 𝐾a1=2.350, p 𝐾a2 = 9.778 ) to adjust the pH to 2.75?
- hcbiochemLv 73 weeks ago
In glycine hydrochloride, both the amino group and the carboxyl group are protonated.
Relevant Equilibrium: GlyH <--> H+ + Gly-
(I am completely ignoring the second ionization because the only important one here is the ionization of the carboxyl group.)Use the Henderson-Hasselbalch equation to calculate the ratio of protonated and unprotonated glycine at this pH.
pH = pKa + log [Gly-]/[GlyH]
2.75 = 2.35 + log [Gly-]/[GlyH]
[Gly-]/[GlyH] = 2.51
Rearranging gives [Gly-] = 2.51 [GlyH]
In the solution, [Gly-] + [GlyH] = 0.017 M.
Substituting the previous equation into this one gives:
2.51[GlyH] + [GlyH] = 0.017
[GlyH] = 4.8X10^-3 M
[Gly-] = 0.017 - 4.8X10^-3 = 0.012 M
Now, the moles of NaOH added will equal moles of Gly- in the solution.
Moles Gly- = 0.012 mol/L X 0.135 L = 1.6X10^-3 mol = moles NaOH added
Volume NaOH = 1.6X10^-3 mol / 0.100 mol/L = 0.016 L = 16 mL