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3 years ago

In the process of either _ or _ negatively charged _ the reacting atoms form _

Chemistry

1 answer:

yarga [219]3 years ago

4 0

In the process of either gaining or losing negatively charged electrons the reacting atoms form ions.

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3 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

Fed [463]

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$

$4.2 + log\frac{0.05}{0.045} = 4.245$

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows

$pH = pKa + log\frac{base}{acid}$

$4.245 = 3.75 + log\frac{base}{acid}$

$log\frac{base}{acid} = 0.5$

$\frac{base}{acid} = 3.162$

Now we must solve the equation above. This will be done using the following values

$\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)} = 0.464 L$

With these values, we can calculate the volumes of NaOH and HCOOH needed to make the buffer.

NaOH volume

( 0.5 - 0.464)L

0.036L .................... 36ml

HCOOH volume

500 - 36 = 464mL

Learn more about benzoic acid

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3 0

2 years ago

The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O represents an acid-base tritration. The equivalence point occurs when 24.75 mL o

artcher [175]

            moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
            moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
            [H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is [H2SO4] = 0.07729 M

6 0

3 years ago

In the process of either ___ or ___ negatively charged ___ the reacting atoms form ___

In the process of either _ or _ negatively charged _ the reacting atoms form _