How can you calculate pka valve if ph is given?
The pKA of enzyme affects its ionization which could alter enzyme activity. For pH < pKa, the value of vmax is constant and that for pH > pKa, vmax decreases; ie. enzyme activity starts to decline.
By following + or - 2 pH to the pKa value
Each buffer will only be of sufficient capacity within a pH interval of pKa+1 > pH > pKa-1 so the optimal buffer pH-range is maximum 2 units.
Phenols are acidic substances (pH over 7).
How can you calculate pka valve if ph is given?
The pKA of enzyme affects its ionization which could alter enzyme activity. For pH < pKa, the value of vmax is constant and that for pH > pKa, vmax decreases; ie. enzyme activity starts to decline.
the pH of ethanol can be calculated using its pKa value (pKa 15.9) and the Henderson-Hasselbalch equation. pH = pKa - log [AH/A] where [AH/A] the ratio of disassociated versus undisassociated species in solution.
pKa of drug can be determined from Handerson Hasselbatch equation., when conc. of salt become equal to acid i.e. at equivalence point when pH=pka then in H-H equn, pH=pka
pH = pKa + log([A-]/[HA])
This question does not make very much sense but it will somewhat be answered. PH is the measurement of a concentration of hydronium ions in a solution. PKA is the measurement of how much is available. If the concentration and pka of a substance is known, the pH can be calculated.
When the concentration of the weak acid is the same as the concentration of the conjugate base, the pH will be equal to the pKa of the weak acid. This is because from the Henderson Hasselbalch eq:pH = pKa + log [acid]/[conj.base] and when [acid] = [conj.base], the log of 1 = 0 and pH = pKa.
Ka = [H+][A-] / [HA] Hence [H+] = Ka[HA] / [A-] Remember pH = -log(10)[H+] 'logging' both sides. -log(10)[H+] = - log(10)Ka[HA] / [A-] By algebraic manipulation of log. pH = -log[A-]^-1 - logKa - log[HA] pH = log[A-] - logKa - log[HA] pH = pKa - log[HA]/[A-]
pKa approximently 17
By following + or - 2 pH to the pKa value
pKa= pH - log(A/HA) to clarify -log is subtract log E.g A buffer is prepared by adding .15 M of NaOH and .1 of a weak acid, HA. If the pH of the buffer is 8.15, what is the pKa of the acid? pH= 8.15 - log .15/.1 = 7.97
pH = pKa + log([A-]/[HA]) pH = pKa+log([conjugate base]/[undissociated acid]) The actual meaning of pKa: the negative log of the dissociation constant, which is a measure of strength of an acid/base when pKa = pH, there is equal concentration of acid and its conjugate base. pKa helps to understand the nature of acid and base like pH: pKa 2 but 7 but < 10 -- weak base pKa >10 --strong base