Phosphoric acid is triprotic acid (three ionizable hydrogens). The values of its stepwise ionization constants are Ka1= 7.5E-5, Ka2= 6.2E-8, and Ka3= 4.2E-13
1) Write the chemical equation for the first ionization reaction of phosphoric acid with water
2)Write the equilibrium constant expression (Ka1) for this reaction.
3)What would be the pH of a solution when [H3PO4]= [H2PO4-]?
(Note: pH= -log[H3O+])
You know acid + water ionizes to give H3O^+ so take the H3O^+ out and see what's left.
H3PO4 + H2O ==> H3O^+ + H2PO4^-
Then Ka1 = (H^+)(H2PO4^-)/(H3PO4)
For the pH.
The pH when (H2PO4^-) = (H3PO4). Note one is in the numerator, the other is in the denominator, they cancel and you are left with Ka1 = (H^+), convert that to pH.
thanx very much!!!!!
1) The chemical equation for the first ionization reaction of phosphoric acid (H3PO4) with water can be written as:
H3PO4 + H2O ⇌ H3O+ + H2PO4-
2) The equilibrium constant expression (Ka1) for this reaction is obtained by writing an equation using the concentrations of the products and reactants:
Ka1 = [H3O+][H2PO4-] / [H3PO4]
3) To find the pH of a solution when [H3PO4] = [H2PO4-], we can assume that the concentration of water (H2O) remains constant, so we can ignore it in the equation.
Using the equation from step 2 and substituting the change in concentration for [H3PO4] and [H2PO4-] with [H+], we have:
Ka1 = [H+][H+]/[H3PO4]
Since [H3PO4] = [H2PO4-], we can substitute [H2PO4-] with [H+], giving us:
Ka1 = [H+]^2 / [H3PO4]
To find the pH, we need to solve for [H+]. Rearranging the equation gives us:
[H+]^2 = Ka1 * [H3PO4]
[H+] = √(Ka1 * [H3PO4])
Taking the negative logarithm of [H+] gives us the pH:
pH = -log[H+] = -log(√(Ka1 * [H3PO4]))
Substituting the given value of Ka1 and the concentration of H3PO4 in the above equation will give us the pH of the solution.
I don't understand how to do the first one.
I really don't know how to do reactions