Fast pls i will give u like for sure solve this question correctly in 5 min pls A student analyzed a CaCO3 antacid tablet by neutralizing the base with an excess of .60 M HCL and back-titrating the excess with .50M NaOH. He recorded the following data: mass of tablet = 1.3g amount of active ingredient CaCO3 per label = 500 mg volume of HCl added: 25 mL initial buret reading : 34 mL final buret reading : 47.4 mL If the concentration of the provided HCl is actually higher than .60M, would the determined mass of CaCO3 in the antacid tablet be too high or too low?
Fast pls i will give u like for sure solve this question correctly in 5 min pls A student analyzed a CaCO3 antacid tablet by neutralizing the base with an excess of .60 M HCL and back-titrating the excess with .50M NaOH. He recorded the following data: mass of tablet = 1.3g amount of active ingredient CaCO3 per label = 500 mg volume of HCl added: 25 mL initial buret reading : 34 mL final buret reading : 47.4 mL If the concentration of the provided HCl is actually higher than .60M, would the determined mass of CaCO3 in the antacid tablet be too high or too low?
Chapter5: Equilibrium, Activity And Solving Equations
Section: Chapter Questions
Problem 3P
Related questions
Question
Fast pls i will give u like for sure solve this question correctly in 5 min pls
A student analyzed a CaCO3 antacid tablet by neutralizing the base with an excess of .60 M HCL and back-titrating the excess with .50M NaOH.
He recorded the following data:
mass of tablet = 1.3g
amount of active ingredient CaCO3 per label = 500 mg
volume of HCl added: 25 mL
initial buret reading : 34 mL
final buret reading : 47.4 mL
If the concentration of the provided HCl is actually higher than .60M, would the determined mass of CaCO3 in the antacid tablet be too high or too low?
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you