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asambeis [7]
4 years ago
8

The acidic substance in vinegar is acetic acid, HC2H3O2. When 4.00 g of a certain vinegar sample was titrated with 0.200 M NaOH,

25.56 mL of base had to be added to reach the equivalence point. What percent by mass of this sample of vinegar is acetic acid?
Chemistry
1 answer:
mihalych1998 [28]4 years ago
5 0

Answer:

7.7%

Explanation:

the balanced equation for the neutralization reaction is:

CH3COOH+NaOH→CH3COONa+H2O

from the balanced equation above,the mole ratio of CH3COOH to NaOH is 1:1

no of mole of CH3COOH= no of mole of NaOH

no of mole of NaOH= concentration(M) × Volume(L)

=0.2 M× 0.02556=0.005112 mol

since, no of mole of CH3COOH= no of mole of NaOH

no of mole of CH3COOH=0.005112 mol

However ,no of mole =mass/molar mass

molar mass of CH3COOH=60.052g/mol

mass of CH3COOH=0.005112×60.052

=0.307g

percentage by mass of acetic acid in the vinegar=0.307/4

=0.07675×100%

=7.675%

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Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y
d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = 10^{-3.5} = 3.162 * 10^{-4}

From the Ice table

3.162 * 10^{-4} = \frac{x + H^+}{[aspirin]}  = \frac{x^{2} }{0.012-x}

given that the value of Ka is small we will ignore -x

x² = 3.162 * 10^{-4} * 0.012

x = 1.948 * 10^{-3}  

Therefore

[ H⁺ ] = 1.948 * 10^{-3}

given that

pH = - Log [ H⁺ ]

     = - ( -3 + log 1.948 )

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Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is :  2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0
2 years ago
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\Delta S=\frac{Q}{T}

The energy balance:

\delta U=[tex]\delta Q- \delta W

If the process is isothermical the U doesn't change:

0=[tex]\delta Q- \delta W

\delta Q= \delta W

Q= W

The work:

W=\int_{V1}^{V2}P*dV

If it is an ideal gas:

P=\frac{n*R*T}{V}

W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV

Solving:

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Final answer:

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