Acidity/Basicity of 2.75 g MgO + 70 mL 2.4 mol/L HNO3: Find pH

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The discussion focuses on determining the acidity or basicity of a solution formed by adding 2.75 g of MgO to 70 mL of 2.4 mol/L HNO3. Participants highlight the importance of identifying the limiting reagent to find the concentration of ions responsible for the solution's pH. The initial calculations of moles for both MgO and HNO3 are discussed, with confusion arising from differing results compared to the textbook's answer. It is emphasized that the remaining reactants after the reaction dictate the final pH. The conversation suggests a need for clearer stoichiometric calculations to resolve the discrepancies.
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Homework Statement




2.75 g of mgo(s) is added to 70.0 ml of 2.4 mol/l hno3(Aq) is the concentration that results from the solution acidic or basic? What is the concentration of the ion that is responsible for the character of the solution. What is the pH ?

Homework Equations




Mgo+2hno3=mg(no3)2+h20

Ph=-log(h30+)

The Attempt at a Solution


First I found the mols of mgo and the mols of Hno3.
( convert g to mol for Mgo and multiplied the concentration by vol for HNo3)

Since they are not in a 1:1 ratio as seen by the balanced chemical equation I took the one with the greater mol(0.06822) which would be HNo3 and multiplied it by 2(molar ratio)

I then used this mol and divided it by the vol of the solution to get 1.95mol/l
However the book is saying .46.. What did I do wrong?
 
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Hard to say what you did, as your description is rather cryptic.

Hint: this is a limiting reagent question. Whatever is left after the reaction will be responsible for the solution pH.

Please pay attention to the way you type formulas. No such things as mgo or hno3.
 
Borek said:
Hard to say what you did, as your description is rather cryptic.

Hint: this is a limiting reagent question. Whatever is left after the reaction will be responsible for the solution pH.

Please pay attention to the way you type formulas. No such things as mgo or hno3.


Usually you use the excess to find the concentration of the ion. What would the limiting reagent have to do with it?
 
I know how you are supposed to do this problem however I am not getting the answer
 
Ignore the pH part now. Solve the stoichiometry. What is left after the reaction?
 
Isn't that what I did above?
 
I multiplied by 2 using stoichmetry
 
I got it
 
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