Calculate the volume required for complete neutralization of 60.00mL of 0.1500 M KOH(aq).

Question

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.

DrBob222 · Answer

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.

tom · Answer

is this way called unit analysis

Anonymous · Answer

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.