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

`V_(base)=9.442mL`

Step-by-step 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)=(M_(acid)V_(acid))/(M_(base))=(0.1894M*10.00mL)/(0.2006M)\\\\ V_(base)=9.442mL`

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