A sample containing HCl is titrated with NaOH: HCl + NaOH à H2O + NaCl A student transfers 25.00 mL of HCl with a volume

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899603
Joined: Mon Aug 02, 2021 8:13 am

A sample containing HCl is titrated with NaOH: HCl + NaOH à H2O + NaCl A student transfers 25.00 mL of HCl with a volume

Post by answerhappygod »

A sample containing HCl is titrated with NaOH:
HCl + NaOH à H2O + NaCl
A student transfers 25.00 mL of HCl with a volumetric pipet toan Erlenmeyer flask, adds phenolphthalein, and titrates with NaOH(0.1000 ± 0.0004M) using a buret, to the endpoint, which requires30.98 mL of NaOH.
Volume of HCl = 25.00 mL ±
Molarity of NaOH = 0.1000 M±
Volume NaOH = 30.98 mL ±
Moles NaOH = ±
Moles HCl = ±
Molarity HCl = ±
Relative uncertainty (%) =
Part 2b:
A student transfers 25.00 mL of HCl with a graduated cylinder toan Erlenmeyer flask, adds phenolphthalein, and titrates with NaOH(0.1000 ± 0.0004M) using a buret, to the endpoint, which requires30.9 mL of NaOH. (The student only reports value of titrant to onedecimal place)
Volume of HCl = 25.00 mL ±
Molarity of NaOH = 0.1000 M ±
Volume NaOH = 30.9 mL ±
Moles NaOH = ±
Moles HCl = ±
Molarity HCl = ±
Relative uncertainty (%)
Part 3: Multiple trials
Given the following data, calculate the average, averagedeviation and percent deviation using two trials and again forthree trials:
Molarity NaOH
Trial 1
0.1034
Trial 2
0.1056
Trial 3
0.1045
Average of Trial 1+2:_____________
Average Deviation Trial 1+2: ____________
Percent Deviation Trial 1+2: ____________
Average of Trial 1-3:_____________
Average Deviation Trial 1-3: ____________
Percent Deviation Trial 1-3: ____________
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply