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Nana76 [90]
3 years ago
8

Explain why the coefficients on the left side of the equation don’t necessarily equal the coefficients on the right side of the

equation. Is this possible if mass is being conserved?
Chemistry
1 answer:
Eduardwww [97]3 years ago
6 0

Answer:

Explanation:

Short answer: Yes.

The coefficients may not be conserved, but mass always has to be. Take this equation as an example

2 Mg3P2 ===>   6Mg + P4

There is a 2 on the left side and 6 and 1 on the right. I hope you mean that the coefficient 2 is not equal to 7.

But let's look a little closer. You have to look at the molecular structure of the left and right side.

2Mg3P2 has 6 Mgs and 4 Ps on the left side.

6Mg is on the right. They are free standing.

P4    has 1 molecule consisting of 4 Ps.

Everything balances.

This is a terrific question to be asking. You need to understand the internal balance numbers vs the molecular ones on the out side.

That sounds like a bit of gobbledygook and it takes a bit of study.

2 Mg3P2 means that Mg3P2 is written twice.

Mg3P2 ==> "3 Mg2+   +  2P3+   and there is another one written the same way.

Mg3P2 ==> "3 Mg2+   +  2P3+  

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A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to
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Explanation:

It is known that pK_{a} of propionic acid = 4.87

And, initial concentration of  propionic acid = \frac{0.19}{1.20}

                                                                       = 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

                                                             = 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence,      pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.216}{0.158}

                        = 4.87 + log (1.36)

                        = 5.00

  • Therefore, when 0.02 mol NaOH is added  then,

     Moles of propionic acid = 0.19 - 0.02

                                              = 0.17 mol

Hence, concentration of propionic acid = \frac{0.17}{1.20 L}

                                                                 = 0.14 M

and,      moles of sodium propionic acid = (0.26 + 0.02) mol

                                                                  = 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.28 mol}{1.20 L}

                           = 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.23}{0.14}

                        = 4.87 + log (1.64)

                        = 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

  • Therefore, when 0.02 mol HI is added  then,

     Moles of propionic acid = 0.19 + 0.02

                                              = 0.21 mol

Hence, concentration of propionic acid = \frac{0.21}{1.20 L}

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Hence, concentration of sodium propionic acid will be calculated as follows.

                        \frac{0.24 mol}{1.20 L}

                           = 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

             pH = pK_{a} + log(\frac{[salt]}{[acid]})

                       = 4.87 + log \frac{0.2}{0.175}

                        = 4.87 + log (0.114)

                        = 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.

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