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Law Incorporation [45]
3 years ago
12

You are given sodium acetate, 1m hcl, nahco3 and na2co3. Determine which of these four you would need and then show calculations

to make buffer ph
Chemistry
1 answer:
valentina_108 [34]3 years ago
5 0

solution:

the given compounds are sodium acetate, 1M HCL,NaHCO₃ and Na2CO₃

pH of the buffer solution is 4.7

the value of pKa of sodium bicarbonate is 6.37

the value of pKa of acetic acid is 4.7

calculate concentration of acetic acid by using the following forumula

pH=pKa+lag[salt]/[acid]

substitute the pH and Pka values in the formula.

4.7=4.7+log[salt]/[acid]

log[salt]/[acid]=0

thus, the concentration ratio of the salt and acid should be equal to each other.

Thus, concentration of sodium acetate is 0.05M

Concentration of sodium acetate= concentration of acid

= 0.05M

Volume of the buffer solution is 100mL

The buffer solution can be prepared as 0.05M of 50mL sodium acetate will react with 0.05M of 50mL of 0.05M of HCL.

The chemical equation for neutralization of the weak base with strong can be represented as show as

CH₃COONa+HCL-->CH₃COOH+NaCL


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0.0277 M.

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How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose
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Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

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Mole of H3PO4 = 0.3 × 0.2

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Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

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Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

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