Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 22 mL of 0.22 M H₃BO₃(aq) with 0.21 M NaOH(aq). The K_a of H₃BO₃ is 5.8 x 10^-10

^{Answer is 11.13}

Answer 1

Solution :-

lets calculate the moles of the acid

moles = molarity * volume

         = 0.22 mol per L * 0.022 L

        = 0.00484 mol

now lets calculate the volume of the base

volume = moles / molarity

          = 0.00484 mol / 0.21 mol per L

         = 0.023 L

0.023 L * 1000 ml / 1 L = 23.0 ml

so the total volume at equivalence point = 22 ml + 23 ml = 45 ml

at equivalence point all acid is congverted to conjugate base

therefore the concentration of the conjugate base at the equivalenc epoint is

0.00484 mol / 0.045 L = 0.10756 M

now lets calculate the concnetration of the OH- when conjugate base dissociates in water

H2BO3- + H2O   ----- > H3BO3 + OH-

0.10756 M                     0              0

-x                                +x             +x

0.10756 -x                     x             x

need to use the kb

kb = kw / ka

kb = 1*106-14 / 5.8*10^-10

kb = 1.724*10^-5

kb = [H3BO3][OH-]/[H2BO3-]

1.724*10^-5 = [x][x]/[0.10756-x]

since kb is very small we can neglect the x from the denominator

then we get

1.724*10^-5 = [x][x]/[0.10756]

1.724*10^-5 * 0.10756 = x^2

1.854*10^-6 = x^2

taking square root of both sides we get

0.00136 = x = [OH-]

now lets calculate the pOH

pOH= -log [OH-]

pOH= -log [0.00136]

pOH = 2.87

pOH+ pH = 14

pH= 14 - pOH

pH= 14-2.87

pH= 11.13