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Posted by on Saturday, September 29, 2012 at 5:04am.

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! :)

  • Chemistry - , Saturday, September 29, 2012 at 2:38pm

    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^+)

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