1) Determine a pH at which pH more than 99% of CH3COOH will be in a form that po
ID: 1000510 • Letter: 1
Question
1) Determine a pH at which pH more than 99% of CH3COOH will be in a form that possesses a charge. (The pKa of CH3COOH is 4.76)2) Determine a pH at which pH more than 99% of CH3NH3+ will be in a form that possesses a charge. (The pKa of CH3NH3+ is 10.7)
1) Determine a pH at which pH more than 99% of CH3COOH will be in a form that possesses a charge. (The pKa of CH3COOH is 4.76)
2) Determine a pH at which pH more than 99% of CH3NH3+ will be in a form that possesses a charge. (The pKa of CH3NH3+ is 10.7)
2) Determine a pH at which pH more than 99% of CH3NH3+ will be in a form that possesses a charge. (The pKa of CH3NH3+ is 10.7)
Explanation / Answer
1)The relationship between pKa and pH is mathematically represented by Henderson-Hasselbach equation:
pH = pKa + log [A-] / [HA]
where [A-] represents the deprotonated form and [HA] represents the protonated form.
In acetic acid: CH3COOH <===> CH3COO- + H+ .
pH = pKa + log [CH3COO-] / [CH3COOH ]
If we want a pH at which more than 99% of CH3COOH will be in a form that possesses a charge: CH3COO-
pH = 4.76 + log [99] / [1 ] = 6.76
2) It is the same case.The reaction (as an acid) is CH3NH3+ <===> CH3NH2+ H+
pH = pKa + log [CH3NH2] / [CH3NH3+] Notice that CH3NH3+ is the form that possesses a charge.
pH = 10.7 + log [1] / [99] = 8.70 .
You can solve it as base as follows:
Methyl amine CH3NH2 is a base: CH3NH2 + H2O <===> CH3NH3+ + OH-
And Henderson-Hasselbach equation:
pOH = pKb + log [HB+] / [B] =pKb + log [CH3NH3+] / [CH3NH2]
pOH = pKb + log [CH3NH3+] / [CH3NH2]
We have pKa=10.7 and we can calculate pKb = 14-pKa = 14-10.7 = 3.3
Then, as CH3NH3+ is the form that possesses a charge:
pOH = pKb + log [CH3NH3+] / [CH3NH2] = 3.3 + log [99] / [1] = 5.30
Finally pH = 14 - pOH = 14 - 5.30 = 8.70
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