Please help solve I made a solution of NaOH with a concentration of 0.1M that is
ID: 1022147 • Letter: P
Question
Please help solve
I made a solution of NaOH with a concentration of 0.1M that is used to titrate a solution of HCl with a concentration of 0.1M
Standardization of a Sodium Hydroxide Solution
Tabulate your results as shown below (also see in the examples under the heading “Sample calculation”).
Trial 1
Trial 2
Trial 3
Initial Burette reading (mL)
17.3 mL
17.5 mL
18.2 mL
Final burette reading (mL)
24.0 mL
24.3 mL
25.1 mL
Volume of titrant (NaOH)
6.7 mL
6.8 mL
6.9 mL
Average = 6.8 mL
g = 0.1mol/L * 40.00g/mol NaOH * 0.3mL = 1.2g NaOH
Given: HCL Concentration = 0.1 M
Estimate the approximate concentration of the sodium hydroxide solution from the mass of sodium hydroxide and the volume of water used. This concentration is just approximate, and is not to be used in subsequent calculations. However, it may be used as a rough to check to the answer obtained in step 7, below.
Concentration = 1.2g NaOH /(0.3mL *40g/mol NaOH) = 0.1 M NaOH
Write the equation for the reaction between hydrochloric acid and sodium hydroxide.
HCl (aq) + NaOH (aq) è NaCl (aq) + H2O (l)
From the volume of the hydrochloric acid aliquot (the precise volume delivered by your burette, as determined in Experiment A1), and the known concentration, calculate the number of moles of acid that reacted in the neutralization reaction. -5mL HCL of base was used for the titration
Use the equation for the reaction to calculate the number of moles of base that are used in the neutralization reaction.
?
Calculate the average volume of sodium hydroxide used. Be sure to explain why you may have chosen to disregard certain results when computing this average.
Average = (6.7+6.8+6.9)/3 = 6.8mL. Dont know why i may have chosen to disregard certain results.
From the results obtained in steps 5 and 6, determine the concentration of the sodium hydroxide solution.
Trial 1
Trial 2
Trial 3
Initial Burette reading (mL)
17.3 mL
17.5 mL
18.2 mL
Final burette reading (mL)
24.0 mL
24.3 mL
25.1 mL
Volume of titrant (NaOH)
6.7 mL
6.8 mL
6.9 mL
Explanation / Answer
5.
from the reaction it is very clear for 0.1 M of NaOH there should be 0.1M of HCL for neutralization
no of moles of HCl/volume of HCl solution = concentration of the solution'
no of moles of HCl = 0.1*5 = 0.5 moles since from the reaction equal no of moles of base are required 0.5 moles of base are required
6.
when most of the results are around 6.8 as you can clearly see you need to disregard the others as some faulty readings can lead to irregular mean(average) and the volume cannot be measured correctly.
7.
concentration=No of moles/volume = 0.5/6.8 = 0.0735 M
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