Information Sodium carbonate Na 2 CO 3 is a salt. If we look at its ions (Na + a
ID: 700114 • Letter: I
Question
Information
Sodium carbonate Na2CO3 is a salt. If we look at its ions (Na+ and CO32-) we can know it comes from the neutralization reaction between NaOH (a strong base) and H2CO3 (a weak acid):
2NaOH (aq) + H2CO3 (aq) = Na2CO3 (aq) + 2H2O (l)
So we would expect the salt to be a basic salt. Thus, we will calculate its OH- concentration first, using the quadratic formula:
(OH-) = (-Kb + (Kb2 + 4CKb)) / 2
Where Kb = Kw/Ka and C = 0.010 M.
Calculate Kb first:
Kb = Kw / Ka
Kb = 1 x 10-14 / 4.8 x 10-11
Kb = 2.083 x 10-4
Substitute Kb and C in the formula:
(OH-) = (-Kb + (Kb2 + 4CKb)) / 2
(OH-) = (-2.083 x 10-4 + (4.340 x 10-8+ 8.332 x 10-6)) / 2
(OH-) = 1.551 x 10-3 mol/L
Calculate the pOH using the formula:
pOH = -log (OH-)
pOH = -log (1.551 x 10-3 M)
pOH = 2.809
Calculate pH using the formula:
pH = 14 - pOH
pH = 14 - 2.809
pH = 11.19
Answers: (OH-) = 1.551 x 10-3 mol/L, pH = 11.19
Explanation / Answer
6. Calculate the OH concentration and pll of a 0.010M aqueous solution of Sodium carbonate, Na Co,. K.- 4.8 x 10 for HCO