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kompoz [17]
3 years ago
8

What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not

e that the concentration and/or pH value may differ from that in the first question. STRATEGY 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic acid), [A–]/[HA]. 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid. Step 1: Rearrange the Henderson Hasselbalch equation to solve for (A-)/(HA) if the solution is at ph 5.0
Chemistry
1 answer:
Leni [432]3 years ago
7 0

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

<em>1,7378 =  [A⁻]/[HA] </em><em>(1)</em>

As concentration of buffer is 0,15M, it is possible to write:

<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.

i hope it helps!

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If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit
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Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

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We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}

10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al

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Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}

0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

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When a teacher dilutes 50 mL of 2.0 M NaOH to 0.50 M, what volume of NaOH results?
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Answer:

= 200 mL

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M1 = 2.0 M

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V2 = M1V1/M2

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3 years ago
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

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% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
4 years ago
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