and in the t amount ofsolute 96 Concentration- amouat of solution X 100 s.u ngth
ID: 588173 • Letter: A
Question
and in the t amount ofsolute 96 Concentration- amouat of solution X 100 s.u ngthe general titrationfo mil ofMiveM ywala late the olarity (Maortheseidi lia (vine i POST-LAB QUESTIONS: Write formula and show all work. 1. How many ml of 0.150M NaOH (base) solution are required to neutralize 35.oml of 0.220M HCI (acid) solution? 2. Titration of a 25.00 mL KOH (base) solution required 200.00 ml of 00050 M acetic acid. What is the molarity of the KOH solution? 3. A 25.00mL sample of an HCI (acid) solution is placed in a flask with a few drops of phenolphthalein indicator. 32.60mL of a 0.185 M NaOH solution is needed to reach the endpoint, what is the concentration of the HCl solution? 4. What is the molarity of a solution prepared by dissolving 100g of Naof in emough water to make 225mL of soluaion 5. What is the % concentration (nv) of a solution prepared by dissolving 10.0g of NaOH in enough water to make 225mL of solution? V026Explanation / Answer
The molarity of one compnent required to neutralize another can be calculated using the formula
M1V1 = M2V2
1) MNaOH x VNaOH = MHCl x VHCl
0.15 M x VNaOH = 0.220 M x 35mL
VNaOH = 51.33 mL
2) MNaOH x VNaOH = MAcOH x VAcOH
MNaOH x 25mL = 0.005 M x 200mL
MNaOH = 0.04 M
3) MNaOH x VNaOH = MHCl x VHCl
0.185 M x 32.6 mL = MHCl x 25 mL
MHCl = 0.24 M
4) Molarity of NaOH solution = (mass/molar mass) x (1000/V(mL))
MNaOH = (10/40)(1000/225) = 1.11 M
5) %(m/V) = (mass of solute/volume of solution) x 100
%(m/V) = 10x100/225 = 4.44%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.