1.48E-11
Fl-
The flouride ion
Potassium fluoride of KF is formed.
The bonding in ammonium boron flouride involves ionic bonds between the positively charged ammonium ion (NH4+) and the negatively charged boron fluoride ion (BF4-), as well as covalent bonds within the boron fluoride ion.
The lithium ion will be stripped of its electron by the flouride ion, resulting in an ionic bond, where the lithium atom will have 0 valence electrons and the fluoride ion will have 8
Calcium Flouride Ca+3F=CaF3 (balanced equation)
Fluoride ion is symbolised by F- .(Mind the spelling of flUOride, it is not flouride)
The Kb value for the conjugate base CN- (cyanide ion) is 2.5 x 10^-5.
To find the pOH from the given pH value (pH = 11.32), subtract the pH from 14 (pH + pOH = 14) to get pOH. Then use the pOH to find the hydroxide ion concentration in the solution. Finally, use the hydroxide ion concentration along with the initial concentration of hypoiodite ion to determine the KB value for the hypoiodite ion using the expression for the KB of a weak base.
The relationship between Ka and Kb values is that they are related by the equation Kw Ka Kb, where Kw is the ion product of water. If you know the Kb value, you can determine the Ka value by rearranging the equation to solve for Ka.
The ammonium ion is NH4+ Fluoride is F- Therefor ammonium fluoride is NH4 F
To find the pH using the base dissociation constant (Kb), you first need to determine the concentration of the base. Then, use the Kb value to calculate the hydroxide ion concentration. Finally, use the hydroxide ion concentration to find the pOH, and then convert it to pH using the formula pH 14 - pOH.