Answer to Question #93244 in General Chemistry for Miguel Lacsson

Question

Assuming a density of 5 percent acetic acid by mass solution is 1.0g/mL, determine the volume of the acetic acid solution necessary neutralize 25.0 mL of .10 NaOH.

Answer 1

CH3COOH + NaOH = CH3COONa + H2On=Cm x V = 10 x 0.025= 0.25 moles of NaOH

And m= n x M = 0.25 x 40 = 10 g, V = m/p = 10 / 1 = 10 ml