Given that 10.00mL of 0.1894M CH3COOH was titrated with 0.2006M NaOH in this experiment, calculate the volume, in mL, of NaOH required

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Question:

Given that 10.00mL of 0.1894M CH3COOH was titrated with 0.2006M NaOH in this experiment, calculate the volume, in mL, of NaOH required

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Answer:[tex]V_{base}=9.442mL[/tex]Explanation:Hello!In this case, when acetic acid is titrated with sodium hydroxide, the following chemical reaction is carried out:[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex]Whereas there is a 1:1 mole ratio between the acid and the base, which means that at the equivalence point we evidence:[tex]n_{acid}=n_{base}[/tex]Which in terms of volumes and concentrations is written as:[tex]M_{acid}V_{acid}=M_{base}V_{base}[/tex]Thus, solving for the required volume of base, we obtain:[tex]V_{base}=\frac{M_{acid}V_{acid}}{M_{base}}=\frac{0.1894M*10.00mL}{0.2006M}\\\\ V_{base}=9.442mL[/tex]Best regards!

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