Question

The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.460 moles of a monoprotic weak acid (Ka = 1.0 × 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?

Calculate the [H ] and pH of a 0.000117 M iodoacetic acid solution. Keep in mind that the Ka of iodoacetic acid is 6.68 × 10-5.(use quadratic formula to solve)

Answer 1

let the monoprotic acid be (HA)

moles of HA= 0.46 moles, the reaction between HA and NaOH is HA+ NaOH------->NaA+ H2O

1 mole of HA reacts with 1 mole of NaOH to reach equivalence point.

At half equivalent point, hence moles of NaOH= moles of HA/2= 0.46/2=0.23

moles of NaA formed =0.23 moles

moles of HA remaining = moles used=moles reacted= 0.46-0.23=0.23

let the volume of solution = V

concentrations : A- (from NaA)= 0.23/V, [HA]= 0.23/V

since pH= pKa+ log [A-]/[HA]

given Ka= 1*10^-5, pKa= 5

pH= 5+ log (0.23/0.23)= 5

2. let the iodoacetic acid C2H3IO2 be represented as HA

Ka=6.8*10-5,

HA undergoes ionization as HA+ H2O-------->H3O+A-

Ka= [H3O+] [A-]/[HA]= 6.8*10^-5

preparing the ICE Table

component HA A- [H3O+]

initial 0.000117 0 0

change -x x x

equilibrium 0.000117-x x x

ka= x²/(0.000117-x)= 6.68*10^-5, x²= 0.000117*6.68*10^-5-6.68*10^-5x

x²+6.68*10^-5x -7.815*10^-9=0

this is quadratic equation whose solution is ( not considering the -ve solution)

x= {-6.68*10^-5+[(6.68*10^-5)²+4*7.815*10^-9]^0.5}/2

x=[H3O+]= 6.11*10^-5, pH= 4.21