1. Acetic Acid is an important component of vinegar. A 10.00-mL sample of vinega

Question

1. Acetic Acid 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 balanced equation for this neutralization reaction. 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?

2. How many milliliters 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

1) Solution

NaOH Na+1 + OH-1
HC2H3O2 + H2O H3O+1 + C2H3O2-1

1 mole of NaOH gives 1 mole of OH-1 ions, and 1 mole of HC2H3O2 gives 1 mole of H3O+1 ions.

Moles = Volume in liters * Molarity
Moles of NaOH = 0.01688 Lr * 0.5052 M/L = 0.008528 mole
Moles of HC2H3O2= .008528 mole

Molarity of the vinegar =0.853M
It means 1 liter of vinegar have 0.853 moles of acetic acid.
Now we know that :Mass = Molarity * molar mass

HC2H3O2 = 24+1 + 3 + 32 = 60 g
Acetic acid has mass = 0.853 * 60 = 51.18 g

Vinegar has mass = 1.006 kg = 1,006 grams

51.18/1006 x 100=5.09%

2) The reaction equation is

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)

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