Table 2-3 Dissociation Constants and pK\'s at 25 °C of Some Acids in Common Labo
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Table 2-3 Dissociation Constants and pK's at 25 °C of Some Acids in Common Laboratory Use as Biochemical Buffers Acid Oxalic acid H PO Citric acid Formic acid Succinic acid Oxalate Acetic acid Citrate Citrate2 Succinate 2-(N-Morpholino)ethanesulfonic acid (MES) Cacodylic acid H,co N-(2-Acetamido)iminodiacetic acid (ADA) Piperazine-N,N-bis(2-ethanesulfonic acid) (PIPES) N-(2-Acetamido)-2-aminoethanesulfonic acid (ACES) H, PO K (M) 5.37 × 10-2 7.08 X 10-3 7.41 × 10-4 1.78 × 10-4 6.17 × 10-5 5.37 × 10-5 1.74 X 10-5 1.74× 10-5 3.98 × 10-6 2.29 × 10-6 8.13 × 10-7 5.37 × 10-7 4.47 × 10-7 2.69 × 10-7 1.74× 10-7 1.58 × 10-7 1.51 × 10-7 1.27 (pK 2.15 (pK,) 3.13 (pK,) 3.75 4.21 (pK) 4.27 (pK2) 4.76 4.76 (pK2) 5.40 (pK3) 5.64 (pK2) 6.09 6.27 6.35 (pK,) 6.57 6.76 6.80 6.82 (pK2) 24 orpholino)propanesulfonic aci N-2-Hydroxyethylpiperazine-N'-2-ethanesulfonic 3.39 × 10-8 7.47 acid (HEPES) N-2-Hydroxyethylpiperazine-N'-3-propanesulfonic 1.10 × 10-8 7.96 acid (HEPPS) N-[Tris(hydroxymethyl)methyllglycine (Tricine) Tris(hydroxymethyl)aminomethane (Tris) Glycylglycine N,N-Bis(2-hydroxyethyl)glycine (Bicine) Boric acid 8.91 × 10-9 8.32× 10-9 5.62 X 10-9 5.50 × 10-9 5.75 × 10-10 5.62 X 10-1 1.66 × 10-10 4.68 × 10-11 7.58 × 10-12 4.17 × 10-13 8.05 8.08 8.25 8.26 9.24 9.25 9.78 10.33 (pK2) 11.12 12.38 (pK3 Glycine HCO3 Piperidine HPOExplanation / Answer
Given data,
[H2PO4-] = 0.08 M (the initial value)
[HPO4-2] = 0.12 M (the initial value)
We can use the Henderson-Hasselbalch equation
and pKa2 to calculate the pH and then [H+] of this solution: (pKa2 =6.8; from liturature)
H3PO4 has three ionisations and each ionisation has different pKa values.
pH = 6.8 + log [HPO42-]/[H2PO4-]
pH = 6.8 + log (0.12/0.08)
= 6.976
[H+] = 10^-6.976
= 1.05 x10^-7 M
Now, we can use the first pKa1, [H2PO4-] and the pH of the solution to calculate the concentration of H3PO4 in the solution:
6.976 = 2 + log [0.08]/[H3PO4]
[H3PO4] = 8.454 x 10^-7 M
From the pH, we can [OH-]:
pH + pOH = 14.00
6.976 + pOH = 14.00
pOH = 7.024
[OH-] = 10^-pOH = 9.46X10^-8 M
with K2HPO4 and KH2PO4 concentrations, we can calculate the [K+]
From K2HPO4, [K+] = 2(0.08) = 0.16 M;
from KH2PO4, [K+] = 0.12 M;
Therefore, total [K+] = 0.28 M
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