Question #15629

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! :)

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! :)

Expert's answer

a.

n of Na2CO3 = 5.3/106 = 0.05 mol

n oh H+ = 0.05*2=0.1mol

M=0.1/2=0.05 mol

pH=-log[H+]=1.3

b.

n of Na2CO3 = 5.3/40 = 0.1325

[H+]=[Na2CO3]

pH = -log[H+]

pH=0.877

n of Na2CO3 = 5.3/106 = 0.05 mol

n oh H+ = 0.05*2=0.1mol

M=0.1/2=0.05 mol

pH=-log[H+]=1.3

b.

n of Na2CO3 = 5.3/40 = 0.1325

[H+]=[Na2CO3]

pH = -log[H+]

pH=0.877

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