## Expert Answer

The dissociates of Carbonic acid, H_{2}CO_{3} is occurs in two steps as follows:.

H_{2}CO_{3} (aq) <=====> H^{+} (aq) + HCO_{3}^{–} (aq) …..(1)

HCO_{3}^{–} (aq) <=====> H^{+} (aq) + CO_{3}^{2-} (aq) ….(2)

Given that;

K_{a1} = 4.2*10^{-7}

And;

pK_{a1} = -log K_{a1}

= -log (4.2*10^{-7})

= 6.38

K_{a2} = 4.8*10^{-11};

pK_{a2} = -log K_{a2}

= -log (4.8*10^{-11})

= 10.32

The pH of buffer is 7.2 hence we choose H_{2}CO_{3}/HCO_{3}^{–} ,because the pK_{a1} (6.38 )of H_{2}CO_{3} is neatest within unit of the pH (7.2).

Now,

Use Henderson-Hasslebach equation as below:

pH = pK_{a1} + log [HCO_{3}^{–}]/[H_{2}CO_{3}]

7.2 = 6.38 + log [HCO_{3}^{–}]/[H_{2}CO_{3}]

0.82 = log [HCO_{3}^{–}]/[H_{2}CO_{3}]

[HCO_{3}^{–}]/[H_{2}CO_{3}] =10^0.82

[HCO_{3}^{–}]/[H_{2}CO_{3}] = 6.61

[HCO_{3}^{–}] = 6.61*[H_{2}CO_{3}] ….(3)

Here given that [H_{2}CO_{3}] = 0.250 M;

therefore, [HCO_{3}^{–}] = 6.61*0.250 M

= 1.65175 M.

Moles of NaHCO_{3} required = molarity * volume in L

=(1.65175 mol/L)*(5 L)

= 8.26 mole (1 M = 1 mol/L).

Mass of NaHCO_{3} required = (8.26 mole)*(84.0 g/mol)

**= 693.8 g**