Question

Despite its formula, boric acid, H3BO3, is a monoprotic acid. A 25.00 mL sample of this acid required 15.66 mL of 0.1724 M NaOH to reach the equivalence point where the pH= 11.13. Determine the original concentration of boric acid and the equilibrium constant, Ka.

Answer 1

H3BO3 + NaOH --- > NaH2BO3 + H2O

Calculate the moles of conjugate base H2BO3-

mol ratio is 1:1

Moles of conjugate base are calculated using moles of base NaOH and moles of base are calculated using given volume and molarity (mol = volume in L x molarity)

Mol of conjugate base = moles of NaOH x 1 mol conjugate base/ 1 mol NaOH

= ( 0.01566L x 0.1724 M ) x 1 mol conjugate base / 1 mol NaOH

0.01566*0.1724=0.0027 mol conjugate base

Calculate concentration of conjugate base

[H2BO3-] = moles of H2BO3- / total volume in L

0.0027/(0.01566+0.025)=0.0664 M

Use ICE table to determine equilibrium concentrations.

We are given pH and pOH is calculated using it.

pOH = 14-pH= 14-11.13=2.87

Calculate hydroxide ion ( OH-) concentration

[OH-]= 10^(-2.87) = 0.00135 M

    H2BO3-(aq) + H2O (l) --- > H3BO3 (aq) + OH-(aq)

I   0.0664                                   0                            0

C -x                                            +x                        +x

E ( 0.0664-x)                             0.00135           0.00135

Value of x is 0.00135 therefore

Equilibrium concentration of conjugate base =   0.00664-0.00135=0.00529 M

Use kb expression

Kb = (0.00135)^2/ 0.00529 = 0.000344

Original concentration of acid = mole / total volume

Mol ratio between conjugate acid and NaOH is 1:1

Therefore moles of acid = moles NaOH x 1mol acid / 1mol NaOH

0.0027 moles of acid

[H3BO3] = 0.0027 / 0.025 L = 0.11 M

We know ka = 1.0 E-14 / kb = 1.0 E-14 / 0.00344 = 0.00344

Ka = 2.9 x 10 ^-11