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.
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.
is this way called unit analysis
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.
To calculate the volume required for complete neutralization using unit analysis, you need to consider the stoichiometry of the reaction and the molar ratios between the reactants. Let's break it down step by step:
Step 1: Write the balanced chemical equation for the neutralization reaction between KOH and HNO3:
KOH + HNO3 → KNO3 + H2O
From the balanced equation above, we can see that the molar ratio between KOH and HNO3 is 1:1, meaning that 1 mole of KOH reacts with 1 mole of HNO3.
Step 2: Calculate the number of moles of KOH:
Given that the initial volume of KOH solution is 60.00 mL (0.06000 L) and the concentration of KOH is 0.1500 M, you can use the formula:
moles of KOH = volume (L) × concentration (M)
moles of KOH = 0.06000 L × 0.1500 M
moles of KOH = 0.009 mol
Step 3: Determine the volume of HNO3 required using the molar ratio:
Since the molar ratio between KOH and HNO3 is 1:1, the volume of HNO3 required will be the same as the volume of KOH used.
Therefore, the volume of HNO3 required for complete neutralization is 0.06000 L (or 60.00 mL).
So, the volume required for complete neutralization of 60.00 mL of 0.1500 M KOH(aq) with 0.1500 M HNO3(aq) is 0.06000 L.