Acetic acid (HC2H3O2) is an important component of vinegar. A 10.00 ml sample of
ID: 608315 • Letter: A
Question
Acetic acid (HC2H3O2) is an important component of vinegar. A 10.00 ml sample of vinegar is titrated with 0.5052 M NaOH, and 16.88 mL are required to neutralize the acetic acid that is present. a.) Write a balance equation for this neutralization reaction. *HC2H3O2 + NaOH -> HOH + NaC2H3O2 (Is that correct???) b.) What is the molarity of the acetic acid in this vinegar? c.) If the density of the vinegar is 1.006 g/mL, what is the mass percent of acetic acid in the vinegar? How many millimeters of stomach acid (assume it is 0.035 M HCL) can be neutralized by an antiacid tablet containing 0.231 g Mg(OH)2???Explanation / Answer
NaOH ? Na+1 + OH-1 HC2H3O2 + H2O ? H3O+1 + C2H3O2-1 1 mole of NaOH supplies 1 mole of OH-1 ions, and 1 mole of HC2H3O2 supplies 1 mole of H3O+1 ions. Moles = Volume in liters * Molarity Moles of NaOH = 0.01688 Liter * 0.5052 mole/liter = 0.008528 mole Moles of HC2H3O2= .008528 mole The molarity of the vinegar is 0.853M This means 1 liter of vinegar contains 0.853 moles of acetic acid. Mass = Molarity * molar mass HC2H3O2 = 1 + 24 + 3 + 32 = 60 g Mass of acetic acid = 0.853 * 60 = 51.18 g Mass of Vinegar = 1.006 kg = 1,006 grams 51.18/1006 x 100=5.09% How many millimeters of stomach acid (assume it is 0.035 M HCL) can be neutralized by an antiacid tablet containing 0.231 g Mg(OH)2??? Mg(OH)2(aq.) + 2HCl(aq.) --> MgCl2(aq.) + 2H2O(l) n(HCl)/n(Mg(OH)2)= 2/1 n(HCl)= 2*n(Mg(OH)2) c= n/V; n= c*V n= m/M c(HCl)*V(HCl)= 2*m(Mg(OH)2)/M(Mg(OH)2) V(HCl)= 2*m(Mg(OH)2)/(M(Mg(OH)2)*c(HCl))= 2 * 0,231 g / (58,326 g mole^-1 * 0,035 mole dm^-3)= 0,226 dm^3= 226 cm^3 (cm^3= mL)