Answer to Question #96280 in General Chemistry for Boye Zephaniah

Question

Vinegar is approximately 4.5%(v/v) aqueous acetic acid [sp. gr. 1.05, purity=99.6% w/w]. How much NaOH would have to be added to 100mL of vinegar to completely neutralize the acid?

Answer 1

V(vinegar)=100 mL

d(AcOH)=1.05 g/mL

x=0.045

$V(AcOH)=x*V(vinegar)$

$m(AcOH)=d(AcOH)*V(AcOH)=d(AcOH)*x*V(vinegar)$

$n(AcOH)=m(AcOH)/M(AcOH)=d(AcOH)*x*V(vinegar)/M(AcOH)$

$m(NaOH)=d(AcOH)*x*V(vinegar)*M(NaOH)/M(AcOH)$ $m(NaOH)=n(AcOH)*M(NaOH)$ $n(NaOH)=m(NaOH)/M(NaOH)=n(AcOH)$

m(NaOH)=d(AcOH)*x*V(vinegar)*M(NaOH)/M(AcOH)

m(NaOH)=1.05 (g/ml)*0.045*100 (ml)*40(g/mol)/ 60 (g/mol) = 3.15 g