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DanielleElmas [232]
1 year ago
8

A 25. 00 ml sample of acetic acid containing phenolphthalein indicator is titrated with 0. 1067 m naoh. The solution changes col

or after 30. 07 ml naoh has been added. What is the concentration of the acetic acid before titration?.
Chemistry
1 answer:
sasho [114]1 year ago
7 0

The concentration of acetic acid:

The concentration of the acetic acid before titration is 0.128 M

What is titration?

Titration is a quantitative analytical procedure that works by allowing a known analyte to gradually react with a titrant until an endpoint is reached.

Titration for weak acid and strong base:

Moles of acetic acid =  moles of NaOH

Given:

Concentration of NaOH = 0.1067 M

Volume of NaOH = 30.07 ml = 0.03007 L

Calculation:

So, by using the formula, Concentration = Moles/Volume

Moles of NaOH = concentration x volume = 0.1067 x 0.03007 = 0.0032

Therefore, the moles of acetic acid  = 0.0032 mole

Now, using the formula again for determining the concentration of acetic acid, we get,

Concentration = Moles/Volume

Concentration of acetic acid =  0.0032/0.025 = 0.128M

Learn more about titration here,

brainly.com/question/2728613

#SPJ4

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0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
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Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

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3 years ago
Which of the following is the best example of kinetic energy being transformed into potential energy?
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Explanation:

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Someone please help! this is the last question<br>I only need help with B.<br><br>​
ludmilkaskok [199]

1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. Na₂CO₃ as a limiting reactant

<h3>Further explanation</h3>

Given

Reaction

2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃

Required

mol ratio

Limiting reactant

Solution

The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)

1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67

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