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 +0  (nbme20#23)

for weak acids pH=pka + log (A-)/(HA) aka henderson hasselbach equation if you solve this, you'll get that pH - pka = log(A-)/(HA) and then again you will get 10 to power ph - pka = A-/HA. [Refer to link in comments] now considering that the ph of our body is 7.3 something, we can solve for pka 10 and pka 6 using this formula. That is for pka 6 and body ph 7.3, it will be 10 to the power of 1.3, would mean that A/HA ratio is a positive number, implying that A- > HA. whereas for when pka is 10, the A-/HA ration is 10 to power of -2.7, meaning that HA>A-.

And as already mentioned in the above comments, we need an iconic form, to eliminate the drug, it would be preferred if the pka is 6 as in that case, A-(ionic) would be in more concentration.

sha2507  https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/buffer-solutions/v/ph-and-pka-relationship-for-buffers refer to this video, to understand it a little better, and for the formula with the table, skip to 7:26




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submitted by sha2507(0),

for weak acids pH=pka + log (A-)/(HA) aka henderson hasselbach equation if you solve this, you'll get that pH - pka = log(A-)/(HA) and then again you will get 10 to power ph - pka = A-/HA. [Refer to link in comments] now considering that the ph of our body is 7.3 something, we can solve for pka 10 and pka 6 using this formula. That is for pka 6 and body ph 7.3, it will be 10 to the power of 1.3, would mean that A/HA ratio is a positive number, implying that A- > HA. whereas for when pka is 10, the A-/HA ration is 10 to power of -2.7, meaning that HA>A-.

And as already mentioned in the above comments, we need an iconic form, to eliminate the drug, it would be preferred if the pka is 6 as in that case, A-(ionic) would be in more concentration.

sha2507  https://www.khanacademy.org/science/chemistry/acid-base-equilibrium/buffer-solutions/v/ph-and-pka-relationship-for-buffers refer to this video, to understand it a little better, and for the formula with the table, skip to 7:26 +