Question

I have had this question in my homework, and honestly I got stuck in these questions. I managed to convert 5.3g of Na2CO3 into moles, but then I got stuck and do not know what to do next to solve the question.

If anyone knows how to work this out, help from anyone is greatly appreciated (:

The addition of 5.3g of anhydrous sodium carbonate to 2.00dm³ of an aqueous solution of a strong acid exactly neutralises the acid

2H+ (aq) + (CO3)2-(aq) ==> CO2 (g) + H2O (l)

a. Calculate the pH of the original solution

b. In a second experiment, 5.3g of NaOH were added, instead of sodium carbonate, to 2.00dm³ of the original solution. Calculate the pH of the resulting solution.

Thanks! :)

Answer 1

a.
mols Na2CO3 = 5.3/106 = ?
mols H^+ initially = ? x 2
M acid = mols/L soln = 2?/2 = x
pH = -log(H^+) = xx

b.
mols NaOH = 5.3/40 = ?
mols H^+ = same
M acid = mols/L soln
pH = -log(H^+)