Question #100673
5. Jacinda is attempting to manufacture some aspirin to get rid of her headache. In it, she needs to adjust the pH, so she makes a weak base to use in the process, so she reacts 125 mL of 5 M acetic acid with enough 2.6 M barium hydroxide to react to completion. How many mL of barium hydroxide are used? What is the final concentration of the salt that is produced from this reaction?
1
Expert's answer
2019-12-22T02:38:21-0500

Solution.

To find the number of ml of barium hydroxide through the law of equivalents, it is necessary to translate all molar concentrations to normal.

Cm(CH3COOH)=Cn(CH3COOH)Cm(CH3COOH) = Cn(CH3COOH)

Cn(Ba(OH)2)=2×2.6=5.2 NCn(Ba(OH)2) = 2 \times 2.6 = 5.2 \ N

Because of this reaction:

2CH3COOH+Ba(OH)2=Ba(CH3COO)2+2H2O2CH3COOH + Ba(OH)2 = Ba(CH3COO)2 + 2H2O

Law of equivalents:

C1×V1=C2×V2C1 \times V1 = C2 \times V2

V(Ba(OH)2)=Cn(CH3COOH)×V(CH3COOH)Cn(Ba(OH)2)V(Ba(OH)2) = \frac{Cn(CH3COOH) \times V(CH3COOH)}{Cn(Ba(OH)2)}

V(Ba(OH)2) = 120.2 ml

Since the law of equivalent quantities of the starting materials are equal, then the amount of substance of salt equal to the amount of substance of a base or acid.

n(Ba(CH3COO)2)=5.2×0.1202=0.63 moln(Ba(CH3COO)2) = 5.2 \times 0.1202 = 0.63 \ mol

The volume of the solution is equal to the sum of the volumes merged into the chemical reactor.

V=V(CH3COOH)+V(Ba(OH)2)V = V(CH3COOH) + V(Ba(OH)2)

V = 245.2 ml

C(Ba(CH3COO)2)=n(Ba(CH3COO)2)VC(Ba(CH3COO)2) = \frac{n(Ba(CH3COO)2)}{V}

Cn(Ba(CH3COO)2) = 2.57 N

Answer:

V(Ba(OH)2) = 120.2 ml

Cn(Ba(CH3COO)2) = 2.57 N


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