A $5.00mL$ volume of acetic acid required a $45mL$ volume of $0.100molL^-1$ $NaOH$ for equivalence. What is the concentration of the acetic acid solution?

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Answer 1 · 2016-11-20T11:36:13

Approx. $1*mol*L^-1$ , with respect to acetic acid.

$"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$ , or $C=n/V$ , and clearly $"concentration"$ has units of $mol*L^-1$ .

We need a reaction:

$H_3C-CO_2H(aq) + NaOH(aq) rarr H_3C-CO_2^(-)""^(+)Na(aq) + H_2O(aq)$

And thus, since
$"moles of sodium hydroxide "-=" moles of acetic acid"$ ,

there were $45.0*mLxx10^-3L*mL^-1xx0.100*mol*L^-1=$ $4.50xx10^-3*mol$ of acetic acid in that original $5*mL$ volume.

Again, $"Concentration"$ $=$ $"Moles of solute"/"Volume of solution"$

$=$ $(4.50xx10^-3*mol)/(5xx10^-3L)$

A 5.00*mL5.00*mL volume of acetic acid required a 45*mL45*mL volume of 0.100*mol*L^-10.100*mol*L^-1 NaOHNaOH for equivalence. What is the concentration of the acetic acid solution?