![From the dissociation constants Ka and Kb for an acid and its conjugate base, show that Ka· Kb = Kw . From the dissociation constants Ka and Kb for an acid and its conjugate base, show that Ka· Kb = Kw .](https://haygot.s3.amazonaws.com/questions/1608485_1740056_ans_85323eb8ce264097bbbec05b4cddfb7e.jpg)
From the dissociation constants Ka and Kb for an acid and its conjugate base, show that Ka· Kb = Kw .
![How is Kw=Ka x Kb if each of those equilibrium constants include the concentration of another substance in them? Kw=H3o x oh but ka which is h3o also includes another substance? : How is Kw=Ka x Kb if each of those equilibrium constants include the concentration of another substance in them? Kw=H3o x oh but ka which is h3o also includes another substance? :](https://preview.redd.it/rcg1yr1sek751.png?width=373&format=png&auto=webp&s=9c1ddb6eb140cc671a776642b7dad30390f9db4f)
How is Kw=Ka x Kb if each of those equilibrium constants include the concentration of another substance in them? Kw=H3o x oh but ka which is h3o also includes another substance? :
![CALCULATING Ka Kb Kw, ICE Tables, ACID DISSOCIATION CONSTANT Power Points & M.C. WITH ANSWERS (59PGS) | Teaching Resources CALCULATING Ka Kb Kw, ICE Tables, ACID DISSOCIATION CONSTANT Power Points & M.C. WITH ANSWERS (59PGS) | Teaching Resources](https://d1e4pidl3fu268.cloudfront.net/ea0ab3df-3e2f-48b1-bd7f-6960ac591bcc/00.jpg)
CALCULATING Ka Kb Kw, ICE Tables, ACID DISSOCIATION CONSTANT Power Points & M.C. WITH ANSWERS (59PGS) | Teaching Resources
What is value of Ka and kb for water (H20) and how numerically prove that ka ×kb=kw in case of water? - Quora
![From the dissociation constants Ka and Kb for an acid and its conjugate base, show that Ka· Kb = Kw . From the dissociation constants Ka and Kb for an acid and its conjugate base, show that Ka· Kb = Kw .](https://haygot.s3.amazonaws.com/questions/1612500_1740151_ans_bdfb34fc466e41398c65a3412f1b4f83.jpg)
From the dissociation constants Ka and Kb for an acid and its conjugate base, show that Ka· Kb = Kw .
![SOLVED: Helpful Equations pH = -log [H2O+] Kw = Ka × Kb [ATJ[H2O+1 [HA]] = pOH -log [OH-] 14 = pH + pOH [HA][OH-] [A-] Kb Henderson-Hasselbalch Equation [A-] pH = pKa log - [HA] SOLVED: Helpful Equations pH = -log [H2O+] Kw = Ka × Kb [ATJ[H2O+1 [HA]] = pOH -log [OH-] 14 = pH + pOH [HA][OH-] [A-] Kb Henderson-Hasselbalch Equation [A-] pH = pKa log - [HA]](https://cdn.numerade.com/ask_images/9f4de4f19628424f8aea53cff3ecadcb.jpg)