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

Answer 1

Answer:

$V_{base}=9.442mL$

Explanation:

Hello!

In this case, when acetic acid is titrated with sodium hydroxide, the following chemical reaction is carried out:

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

Whereas there is a 1:1 mole ratio between the acid and the base, which means that at the equivalence point we evidence:

$n_{acid}=n_{base}$

Which in terms of volumes and concentrations is written as:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, solving for the required volume of base, we obtain:

$V_{base}=\frac{M_{acid}V_{acid}}{M_{base}}=\frac{0.1894M*10.00mL}{0.2006M}\\\\ V_{base}=9.442mL$

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