REPORT FOR EXPERIMENT 22 (continued) NAME QUESTIONS AND PROBLEMS If you had adde
ID: 580567 • Letter: R
Question
REPORT FOR EXPERIMENT 22 (continued) NAME QUESTIONS AND PROBLEMS If you had added 50 mL of water to a sample of KHP instead of 30 mL, would the titra- tion of that sample then have required more, less, or the same amount of base? Explain. 1. 2. A student weighed out 1.106 g of KHP How many moles was that? mol 3. A titration required 18.38 mL of 0.1574 M NaOH solution. How many moles of NaOH were in this volume? mol 4. A student weighed a sample of KHP and found it weighed 1.276 g Titration of this KHP required 19.84 mL of base (NaOH). Calculate the molarity of the base. Forgetful Freddy weighed his KHP sample, but forgot to bring his report sheet along, so he recorded the mass of KHP on a paper towel. During his titration, which required 18.46 mL of base, he spilled some base on his hands. He remembered to wash his hands, but forgot about the data on the towel, and used it to dry his hands. When he went to calculate the molarity of his base, Freddy discovered that he didn't have the mass of his KHP. His kindhearted instructor told Freddy that his base was 0.2987 M. Calculate the mass of Freddy's KHP sample. 5. 6. What mass of solid NaOH would be needed to make 645 ml of Freddy's NaOH solution?Explanation / Answer
1.same amount of base required
Volume of base consumption depands on no of mole of KHP not depands on its concentration, While adding concentration of KHP Changed but not no of mole of KHP
2. No of moles of KHP = Mass of KHP/Molar mass of KHP
Molar mass of KHP = 204.22g/mol
no of mole of KHP = 1.106g/204.22g/mol = 0.005415
3. No of mole of NaOH = (No of mole of NaOH / 1000ml)*Volume of NaOH
=( 0.1574mole/1000ml)*18.38ml
= 0.002893
4. No of mole of KHP = 1.276g/204.22g/mol = 0.006248mole
0.006248mol of KHP represents 0.006248mole of base
volume of base consumed =19.84ml
Molarity of base = (0.006248mol/19.84ml)*1000ml = 0.3149M
5. No of mole of base = (0.2987mol/1000ml)*18.46ml = 0.005514
No of mole of KHP = 0.005514
Mass = No of mole * Molar mass
= 0.005514mole * 204.22g/mole
=1.1261g
6. Molar mass of NaOH = 40g
no of mole in 645ml NaOH solution =( 0.2987mol/1000ml)*645ml = 0.1869mol
mass of Solid NaOH Needed = 40g/mol * 0.1869mol = 7.476g
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.