chemistry

Calculate the volume required for complete neutralization of 60.00mL of 0.1500 M KOH(aq).
a) 0.1500M nitric acid, HNO3.

The answer I got for this question is 0.06 using method c1v1=c2v2 but I want to show using unit analysis. Please help me know how to do this.

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  1. KOH + HNO3 ==> KNO3 + H2O
    mols KOH = M x L = 0.1500 x 0.06000 = 0.009000
    mols HNO3 = mols KOH since 1 mol KOH =- 1 mol HNO3
    Then M HNO3 = mols HNO3/L HNO3 = 0.009/0.1500 = 0.06000 l = 60.00 mL.

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  2. is this way called unit analysis

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  3. Nope, but I don’t think you can use unit analysis to calculate the volume needed to reach the equivalence point. Google unit analysis and see what you come up with.

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