Question

Acetic acid is an important ingredient of vinegar. A sample of 50.0 mL of a commercial vinegar is titrated against a 1.00 M NaOH solution. What is the molarity of acetic acid present in the vinegar if 5.75 mL of the base are needed for the titration? Write out the complete balanced molecular equation with phase labels What is the complete ionic equation? (Remember to check if strong/weak) What is the net ionic equation? Show your work for the calculation of the molarity of HC2H:02 Final Answer

Answer 1

balanced molecular equation
CH3COOH (aq) + NaOH (aq) ---> CH3COONa(aq) + H2O(l)

Ch3COOH -weak acid,NaOH-strong base
split (aq) compounds in to ions we get

Total ionic equation::

CH3COO-(aq) + H+(aq) + Na+(aq) + OH-(aq) ---> CH3COO-(aq) + Na+(aq) + H2O(l)

common ions present in reactants and products side are called spectator ions.Cancel
those ions to get net ionic equation.Here spectator ions are CH3COO-, Na+

Net ionic equation::

H+ (aq) + OH-(aq) ----> H2O(l)
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given
molarity of NaOH M1 = 1M
volume of NaOH, V1 = 5.75 ml
volume of vinegar V2= 50 ml
molarity of vinegar ,M2= ??
Calculate M2 using dilution formula
M1V1= M2V2
M2 = M1V1/2
= 1 M*5.75 ml/50 ml = 0.115 M
Thus molarity of CH3COOH = 0.115 M
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hope it is helpful