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1 year ago

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acetaminophen is a stronger acid than methanol, and sodium methoxide is a stronger base than. write an acid base equilibrium rea

ction in these species (showing structures). In which direction does the equilibrium lie

1 answer:

torisob [31]1 year ago

8 0

Answer:

The equilibrium shifts to the right that is the forward reaction.

Explanation:

The chemical compound known as "Acetaminophen" is a chemical compound that is generally known to a layman as Paracetamol and it belongs to the drug class known as anagelsics which helps in the treatment of pain or say in the reduction of pain. Acetaminophen has the chemical Formula to be C8H9NO2, with the Molar mass of 151.163 g/mol and Boiling point of 420 °C.

The reaction between Acetaminophen and sodium methoxide gives methanol and acetaminophen sodium salt. Therefore, the acid base equilibrium reaction of these species is given as;

C8H9NO2 + CH3ONa <========> CH3OH + acetaminophen sodium salt.

The equilibrium shifts to the right that is the forward reaction.

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1.30mL

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH —> K2SO4 + 2H2O

From the equation above we obtained the following:

Mole of acid (nA) = 1

Mole of base (nB) = 2

The following data were obtained from the question:

Mb = 0.00945M

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Ma = 1.23 × 10^−4M

Using MaVa / MbVb = nA/nB, we can calculate the volume of KOH as illustrated below:

MaVa / MbVb = nA/nB

(1.23 × 10^−4 x 50)/0.00945xVb = 1/2

Cross multiply to express in linear form

1.23 × 10^−4 x 50 x 2 = 0.00945xVb

Divide both side by 0.00945

Vb = (1.23 × 10^−4 x 50 x 2) /0.00945

Vb = 1.30mL

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A 1000.0 ml sample of lake water in titrated using 0.100 ml of a 0.100 M base solution. What is the molarity (M) of the acid in

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The molarity (M) of the acid in the lake water is $0.00001M$ .

<u>Explanation:</u>

In order to estimate the concentration of a solution in molarity, then the total number of moles of the solute is divided by the total volume of the solution.

According to the given information, the formula will be applied for calculating molarity (M) of the acid in the lake water is :

$M_1V_1=M_2V_2$

Here;

$M_1,M_2$ are molarity of acid in the lake water and base solution respectively. $V_1,V_2$ are volume of sample in the lake water and base solution respectively.

Given values are as follows:

$M_1=?\\M_2=0.100M\\V_1=1000ml\\V_2=0.100ml$

Putting these values in above equation :

$M_1V_1=M_2V_2$

$M_1(1000)=(0.100)(0.100)$

$M_1=\frac{(0.100)(0.100)}{1000}$

$M_1=0.00001M$

Therefore, the molarity (M) of the acid in the lake water is $0.00001M$ .

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Hence, we can conclude that most abundant species is $H_{2}PO^{-}_{4}$ and the second most abundant species is $HPO^{2-}_{4}$ .

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