Suppose a student performed a similar standardization titration experiment using the data in the table here....
Suppose a student performed a similar standardization titration experiment using the data in the table here. What is the exact concentration of the NaOH solution? The molar mass of benzoic acid is 122.12 g/mol.
Homework Answers
First, the reaction must be balanced
C6H5COOH + NaOH = NaC6H5COO + H2O
this is 1:1 ratio
Then
mol of NaOH = mol of Benzoic acid
Assume Benzoci acid's mass is correct
then
mol of Benzoic ACid = mass/MW = (0.158) / (122.12 ) = 0.0012938 mol of B.acid
now, this is the actual amount of NaOH that will be neutralized
so
NaOH moles used = 0.0012938 moles of NAOH
Volume of NaOH used to neutralize :
V base = Vfinal - Vinitial = 27.84 mL
Recalculate Molarity of NaOH
[NaOH] = mol/V = (0.0012938) / (27.84*10^-3) = 0.046472 M
roundup
0.046472 -- 0.0465 M
Add Answer to:
Suppose a student performed a similar standardization titration experiment using the data in the table here....
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