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Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution according to the equation H C 2 H 3 O 2 + N a O H ⟶ H 2 O + N a C 2 H 3 O 2 HCX2HX3OX2+NaOH⟶HX2O+NaCX2HX3OX2 If you require 33.98 mL of 0.1656 M N a O H NaOH solution to titrate 10.0 mL of H C 2 H 3 O 2 HCX2HX3OX2 solution, what is the molar concentration of acetic acid in the vinegar? Type answer:

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sebassandin

Answer:

[tex]M_{acid}=0.563M[/tex]

Explanation:

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In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]

Thus, we solve for the molarity of the acid to obtain:

[tex]M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M[/tex]

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