Answer to Question #15629 in Inorganic Chemistry for Matthias
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.
n of Na2CO3 = 5.3/106 = 0.05 mol
n oh H+ = 0.05*2=0.1mol
n of Na2CO3 = 5.3/40 = 0.1325
pH = -log[H+]
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!