Consider the following equation, showing the reaction between dihydrogen phospha

Question

Consider the following equation, showing the reaction between dihydrogen phosphate and water: H:P04-(aq) + H20() HP042-(aq) + H30+(aq) Label the weak acid and the conjugate base in the acid dissociation reaction. And calculate the pKa for the reaction given that the value for Ka for this reaction is 6.32*10~(-8) Write the equilibrium constant expression for the reaction. If you take the logarithm of both sides, multiply both sides by -1, and rearrange, you get the equation below, known as the Henderson- Hasselbalck equation. Suppose you have a beaker containing equal amounts of dihydrogen phosphate and its conjugate base, hydrogen phosphate. Calculate the pH of this solution. a. b. c.

Explanation / Answer

First, let us define Bronsted Lowry acid/base:

Bronsted Lowry acid: any species that will donate H+ (protons) in solution, and makes pH lower (i.e HCl)

Bronsted Lowry base: any species that will accept H+ (protons) in solution, and makes pH higher (NH3 will accept H+ to form NH4+)

Typically, acid/bases are shown in the left (reactants)

when we write the products:

Bronsted Lowery conjugate base = the base formed when the B.L. acid donates its H+ proton ( i.e. HCl -> Cl-

Bronsted Lowery conjugate acid = the acid formed when the B.L. base accept its H+ proton ( i.e. NH4+ has accept H+ proton)

Note that, typically conjugate bases/acids are shown in the right (product) side

So, from your reaction:

H2Y- transforms to H3Y; it is gaining and H+, so it is a BASE

H2Z- transforms to HZ2-; it is losing H+, so it is an ACID

then,

H3Y is the conjugate ACID (since it could donate H+ to form H2Y-)

and HZ2- is the conjugate BASE ( since it could accept H+ to form H2Z-)

left side:

H2PO4- is acting as an acid, H2O acts as a abse

right side

HPO4-2 is acting as a base, H3O+ as ana cid

b)

Ka2 = [HPO4-2][H3O+]/[H2PO4-]

log(Ka2) = [HPO4-2][H3O+]/[H2PO4-])

log(Ka2) = -log([HPO4-2]/[H2PO4-]) + log(H3O+)

multiply by -1

pKa2 = -log([HPO4-2]/[H2PO4-]) + pH

pH = pKa2 + log([HPO4-2]/[H2PO4-])

c.

if H2PO4- = HPO4-2

then

pH = pKa + log(1)

pH = 7.21

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