At first one has to suppose that the concentrations of HA and HB are the same! 1)
Eq.P. at pH=8: You need less OH- (titrant) to reach this, so the acid donated its last proton easier, so it was stronger. (Eq.P. at pH=7 means it isn't weak at all)
1) It is more accurate to compare pH value's half-the-wayof titration, because pH1/2way is exactly equal to pKa value of the acid concerned AND it is independent of concentration: it is buffered!!!
You need to know the volume of the weak acid being titrated so you can find how many moles of base are needed to match that of the acid.
Conjugate base B- is stronger, it has a higher pH at Eq.P.Your first (and also the original) question is answered as follows:At first one has to suppose that the concentrations of HA and HB are the same! 1)Eq.P. at pH=8: You need less OH- (titrant) to reach this, so the acid donated its last proton easier, so it was stronger. (Eq.P. at pH=7 means it isn't weak at all)1) It is more accurate to compare pH value's half-the-wayof titration, because pH1/2way is exactly equal to pKa value of the acid concerned AND it is independent of concentration: it is buffered!!!
Determination of the Dissociation Constant and Molar Mass for a Weak AcidAbstract: We will determine Ka and the molar mass for an unknown weak acid by using a pH meter to record the pH at intervals during the titration with sodium hydroxide. The titration curve and its first derivative will be plotted to establish the equivalence point. Introduction The strength of an acid is defined by its ability to donate a proton to a base. For many common acids, we can quantify acid strength by expressing it as the equilibrium constant for the reaction in which the acid donates a proton to the standard base, water, as shown in the equations below: HA + H2O Û H3O+ + A-, for H3CCOOH: H3CCOOH + H2O Û H3O+ + H3CCOO - The equilibrium constant for a reaction of this type is called the Acid Dissociation Constant, "Ka", for the acid HA Determination of the Dissociation Constant and Molar Mass for a Weak AcidAbstract: We will determine Ka and the molar mass for an unknown weak acid by using a pH meter to record the pH at intervals during the titration with sodium hydroxide. The titration curve and its first derivative will be plotted to establish the equivalence point. Introduction The strength of an acid is defined by its ability to donate a proton to a base. For many common acids, we can quantify acid strength by expressing it as the equilibrium constant for the reaction in which the acid donates a proton to the standard base, water, as shown in the equations below: HA + H2O Û H3O+ + A-, for H3CCOOH: H3CCOOH + H2O Û H3O+ + H3CCOO - The equilibrium constant for a reaction of this type is called the Acid Dissociation Constant, "Ka", for the acid HA Determination of the Dissociation Constant and Molar Mass for a Weak AcidAbstract: We will determine Ka and the molar mass for an unknown weak acid by using a pH meter to record the pH at intervals during the titration with sodium hydroxide. The titration curve and its first derivative will be plotted to establish the equivalence point. Introduction The strength of an acid is defined by its ability to donate a proton to a base. For many common acids, we can quantify acid strength by expressing it as the equilibrium constant for the reaction in which the acid donates a proton to the standard base, water, as shown in the equations below: HA + H2O Û H3O+ + A-, for H3CCOOH: H3CCOOH + H2O Û H3O+ + H3CCOO - The equilibrium constant for a reaction of this type is called the Acid Dissociation Constant, "Ka", for the acid HA
You'd go BOOM! ha ha ha :)
Each cationic acid HA+, when it donates ONE proton H+, will form its conjugated base A of the acid HA+. (Example: NH4+ ammonium, acid NH3 ammonia, base)Each neutral acid HA, when it donates ONE proton H+, will form its conjugated base A- of the acid HA. (Example: CH3COOH acetic acid CH3COO- acetate, base)Each anionic acid HA-, when it donates ONE proton H+, will form its conjugated base A2- of the acid HA-. (Example: HS- (mono)hydrogen sulfide, acid S2- sulfide, base)Each anionic base HA-, when it adopts ONE proton H+, will form its conjugated acid H2A of the base HA-. (Example: HS- mono-hydrogen sulfide, base H2S di-hydrogen sulfide, acid)Each anionic base A-, when it adopts ONE proton H+, will form its conjugate acid HA of base A-. (Example: CH3COO- acetate, base CH3COOH acetic acid)Each neutral base A, when it adopts ONE proton H+, will form its conjugate acid HA of base A-. (Example: NH3 ammonia, base NH4+ ammonium, acid)
no point in trying, their done ha ha ha
For a weak acid, HA...HA ==> H^+ + A^- Ka = [H+][A-]/[HA] Plug these values into the Ka equation. You also must know the [HA] that you start with. Solve for [H+] Take -log [H+] = pH
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weak acid
A strong acid is an acid that ionizes completely in an aqueous solution by losing one proton. The strength of acids can be compared by using pkas. These are found by; For the acid/base reaction - HA A− + H+, hence Ka= [A-][H+]/[HA] pka = -log ka
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-]
parallel lines intersecting lines will meet at some point (ha ha point! get it?)