If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40...
If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate it, what is the acetic acid concentration in the mixture?
Homework Answers
We know the formula
M1V1=M2V2
Given M2=0.258
V2=32.40 ml
V1= 1 ml
Calculate M1=?
Puting value in formula
M1×1 ml = 0.258 molar×32.40 ml
M1 = 8.3592 molar of acetic acid
Add Answer to:
If a 1.00 mL sample of the reaction mixture for the equilibrium
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Calculate the equilibrium constant Kn for the given data . Use the average volume of NaOH rather than doing the calculation for each trial. please show work! A student titrated 5.00 mL of 50,00 mL...
Calculate the equilibrium constant Kn for the given data
. Use the average volume of NaOH rather than doing the calculation
for each trial.
please show work!
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56,57,58,59,60,61,62
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. Use the average volume of NaOH rather than doing the calculation
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please show work!
A student titrated 5.00 mL of 50,00 mL reference solution with 0.698 M NaOH aq) The information of the reference solution is given as Chemical nitia Mols ethyl 0.101 acetato cthanol 0.258 acetic 0.062 acid H20 1.034 HCI 0.066 The titration data for 4 trials is given by A student prepared 4 samples...
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The equilibrium constant for the reaction shown here is K. = 9.0 x 102. A reaction mixture at equilibrium contains [A] = 4.3 x 10-3 M. A (9)=B (9)
The purpose of the lab was to find the equilibrium constant of
the rxn of acetic acid and ethanol to yield ethyl acetate and
water. Hydrochloric acid is used as a catalyst. We have to
determine the equilibrium concentrations of all the species (water
included) present at equilibrium. In test tubes, different
concentrations of the equilibrium system were prepared then
titrated with NaOH to determine the total acid content of the
system at equilibrium. From this you will be able...
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