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What is the pH of a buffer consisting of 0.700 M propionic acid (HC3H5O2, call it HA) and 0.650 M sodium propionate (NaC3H5O2, just call it NaA, with the anion being A)? The Ka for propionic acid is 1.31 x 10-5
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Oreld Hadilberg
What is the pH of a buffer consisting of 0.700 M propionic acid (HC3H5O2, call it HA) and 0.650 M sodium propionate (NaC3H5O2, just call it NaA, with the anion being A)? The Ka for propionic acid is 1.31 x 10-5
Perhaps do your own homework?
These calculations assume that the acid does not dissociate, nor does the propanoate associate with H+ ions. One of these things Is it true true but statistically it makes no odds.
These calculations assume that the acid does not dissociate, nor does the propanoate associate with H+ ions. One of these things Is it true true but statistically it makes no odds.
Perhaps do your own homework?
These calculations assume that the acid does not dissociate, nor does the propanoate associate with H+ ions. One of these things Is it true true but statistically it makes no odds.
kA = [H+][A-]/[HA]
So, [H+] = Ka x [HA]/ [A-]
And pH = - log [H+]
Perhaps do your own homework?
These calculations assume that the acid does not dissociate, nor does the propanoate associate with H+ ions. One of these things Is it true true but statistically it makes no odds.
kA = [H+][A-]/[HA]
So, [H+] = Ka x [HA]/ [A-]
And pH = - log [H+]
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Calculate poH using the Henderson - Hasselbalch equation
pH = pKa + log ([NaA] / [HA] )
pKa = - log 1.31*10^-5 = 4.88
pH = 4.88 + log ( 0.650 / 0.700 )
pH = 4.88 + log 0.928
pH = 4.88 + ( -0.032)
pH = 4.85
Calculate poH using the Henderson - Hasselbalch equation
pH = pKa + log ([NaA] / [HA] )
pKa = - log 1.31*10^-5 = 4.88
pH = 4.88 + log ( 0.650 / 0.700 )
pH = 4.88 + log 0.928
pH = 4.88 + ( -0.032)
pH = 4.85
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