The Ka of sodium hydrogen carbonate is the same as the second ionization constan

Question

The Ka of sodium hydrogen carbonate is the same as the second ionization constant of carbonic acid, which is 4.7 times 10^-11. You want to make 1.0 liters of a pH 10.0 buffer that will resist drastic changes in pH with the addition of 10.0 ml of 1.0 M HCl. You wish to do this by using a mixture of sodium carbonate and sodium hydrogen carbonate. Use the Henderson-Hasselbalch equation to find the molar ratio of sodium carbonate to sodium hydrogen carbonate The number of moles of sodium carbonate should be equal to or greater than the number of moles of acid added. Find the minimum mass of each salt that should be used to make the 1 L solution Calculate pH of your buffer after 5.0 ml of 1.0 M HCl is added.

Explanation / Answer

(a): Required pH of the buffer = 10.0

pKa of the acid, pKa = - logKa = - log( 4.7x10-11) = 10.328

We can calculate the ratio of salt to acid by applying Hendersen equation.

pH = pKa + log[salt] / [acid]

=> 10.0 = 10.328 + log [Na2CO3] / [NaHCO3]

=> log [Na2CO3] / [NaHCO3] = - 0.328

=> [Na2CO3] / [NaHCO3] = 10-0.328 = 0.470 (answer)

(b): Moles of acid that needs to be resisted by the buffer = MxV(L) = 1.0 mol/L x 10 mL x (1L / 1000 mL)

= 0.01 mol

Hence we need to add a minimum of 0.01 mol of Na2CO3

=> minimum mass of Na2CO3 that needs to be added = 0.01 mol x (105.99 g/mol) = 1.06 g

Also

[Na2CO3] / [NaHCO3] = 0.470

=> moles of Na2CO3 / moles of NaHCO3 = 0.470

=> 0.01 mol / moles of NaHCO3 = 0.470

=> moles of NaHCO3 = 0.01 / 0.470 = 0.02128 mol

=> mass of NaHCO3 need to be added = 0.02128 mol x 84.0 g/mol = 1.79 g

Hence we need to add a minimum of 1.06 g Na2CO3 and 1.79 g NaHCO3 (answer)

(c): pH after 5.0 mL of 1.0 M HCl added:

moles of acid added = MxV(L) =1.0 mol/L x 5.0 mL x (1L / 1000 mL) = 0.005 mol

Now 0.005 mol HCl will react with 0.005 mol Na2CO3 to form 0.005 mol NaHCO3.

Hence moles of Na2CO3 remained in solution = 0.01 - 0.005 = 0.005 mol

moles of NaHCO3 remained in solution = 0.02128 mol + 0.005 mol = 0.02628 mol

applying Hendersen equation.

pH = pKa + log [Na2CO3] / [NaHCO3] = pKa + log (moles of Na2CO3 / moles of NaHCO3)

=> pH = 10.328 + log (0.005 mol / 0.02628 mol)

=> pH = 9.61 (answer)

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