A 30.0 mL aliquot of 0.300 M H_3 PO_4 is mixed with 90.0 mL of 0.200 M KOH, and the mixture is evaporated. Will the salt that crystallizes out be K_3 PO_4, K_2 HPO_4, or KH_2 PO_4? Show your calculations by using an ICE table (or two or three). H_3 PO_4(aq) + KOH_(aq) rightarrow KH_2 PO_4(aq) + H_2 O (l) The next proton on the H_2 PO_4^- would start to react only when more OH^- was added, over and above the first mole: KH_2 PO_4(aq) + KOH_(aq) rightarrow K_2 HPO_4(aq) + H_2 O (I) The final proton on the HPO_4^2- would start to react only when more OH^- was added, over and above the second mole: K_2 HPO_4(aq) + KOH_(aq) rightarrow K_3 PO_4(aq) + H_2 O (I)
Ionic Equilibrium
Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibrium is established between the ions and unionized species in a system. Understanding the concept of ionic equilibrium is very important to answer the questions related to certain chemical reactions in chemistry.
Arrhenius Acid
Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
Bronsted Lowry Base In Inorganic Chemistry
Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
A 30.0 mL aliquot of 0.300 M H_3 PO_4 is mixed with 90.0 mL of 0.200 M KOH, and the mixture is evaporated. Will the salt that crystallizes out be K_3 PO_4, K_2 HPO_4, or KH_2 PO_4? Show your calculations by using an ICE table (or two or three).
H_3 PO_4(aq) + KOH_(aq) rightarrow KH_2 PO_4(aq) + H_2 O (l)
The next proton on the H_2 PO_4^- would start to react only when more OH^- was added, over and above the first mole:
KH_2 PO_4(aq) + KOH_(aq) rightarrow K_2 HPO_4(aq) + H_2 O (I)
The final proton on the HPO_4^2- would start to react only when more OH^- was added, over and above the second mole:
K_2 HPO_4(aq) + KOH_(aq) rightarrow K_3 PO_4(aq) + H_2 O (I)
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