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mash [69]
3 years ago
7

g A 500. mL solution contains 0.665 M NaC2H3O2 and 0.475 M HC2H3O2. What mass of HCl in grams needs to be added for the solution

to have a pH of 4.21? The Ka of acetic acid is 1.8 x 10-5.
Chemistry
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

7.38g HCl

Explanation:

Using H-H equation for acetic buffer:

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

<em>Where pKa is -log Ka = 4.74 and [] could be taken as moles of each compound.</em>

The initial moles of each specie is:

[NaC2H3O2]:

0.500L * (0.665mol/L) = 0.3325moles

[HC2H3O2]:

0.500L * (0.475mol/L) = 0.2375 moles

That means total moles are:

[NaC2H3O2] + [HC2H3O2] = 0.57 moles <em>(1)</em>

And solving H-H equation for a pH of 4.21:

4.21 = 4.74 + log [NaC2H3O2] / [HC2H3O2]

0.29512 = [NaC2H3O2] / [HC2H3O2] <em>(2)</em>

Replacing (1) in (2):

0.29512 = 0.57mol - [HC2H3O2] / [HC2H3O2]

0.29512 [HC2H3O2] = 0.57mol - [HC2H3O2]

1.29512 [HC2H3O2] = 0.57mol

[HC2H3O2] = 0.44 moles

The HCl reacts with NaC2H3O2 producing HC2H3O2, that means you need to add:

0.44 moles - 0.2375 moles =

0.2025 moles of HCl

Using molar mass of HCl (36.45g/mol), to convert these moles to grams:

0.2025 moles * (36.45g/mol) =

<h3>7.38g HCl</h3>

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Muriatic acid, HCl, is often used to remove rust. A solution of muriatic acid, HCl, reacts with Fe2O3 deposits on industrial equ
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Answer:

3L

Explanation:

Step 1:

The balanced equation for the reaction.

Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O

Step 2 :

Determination of the masses of HCl and Fe2O3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Fe2O3 = 159.70g/mol

Molar mass of HCl = 36.46 g/mol

Mass of HCl from the balanced equation = 6 x 36.46 = 218.76g

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl

Step 3:

Determination of the mass of HCl needed to react with 439g of Fe2O3. This is illustrated below:

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl.

Therefore, 439g of Fe2O3 will react with = (439 x 218.76) /159.70 = 601.35g of HCl.

Step 4:

Conversion of 601.35g of HCl to mole. This is illustrated below:

Molar mass of HCl = 36.46 g/mol

Mass of HCl = 601.35g

Number of mole = Mass/Molar Mass

Number of mole of HCl = 601.35/36.46

Number of mole of HCl = 16.49 moles

Step 5:

Determination of the volume of the HCl that reacted.

This is illustrated below:

Mole of HCl = 16.49 moles

Molarity of HCl = 5.50 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 16.49/5.5

Volume of HCl = 3L

Therefore the volume of HCl needed for the reaction is 3L

5 0
2 years ago
You have 54.32 grams of PbCl4. How many moles of PbCl4 do you have?
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Answer:

0.156mol

Explanation:

Number of moles of a substance can be calculated from its mass by dividing its mass by molar mass i.e.

Number of moles (n) = mass/molar mass

Molar mass of PbCl4 is as follows, where Pb = 207.2g/mol, Cl = 35.5g/lol

PbCl4 = 207.2 + 35.5(4)

= 207.2 + 142

= 349.2g/mol

Using: mole = mass/molar mass

mole = 54.32 grams ÷ 349.2g/mol

mole = 0.1555

mole = 0.156mol

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Answer:

Five ( 5 ) is the correct answer

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Answer:

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Explanation:

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