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grigory [225]
3 years ago
9

A buffer is prepared by adding 100 mL of 0.50 M sodium hydroxide to 100 mL of 0.75 M propanoic acid. Is this a buffer solution,

and if so, what is its pH?
Chemistry
1 answer:
Yanka [14]3 years ago
7 0
The answer: is yes, It is a buffer solution.

first, we need to get moles of sodium hydroxide and propanoic acid:

moles NaOH = molarity * volume 
                       
                       = 0.5M * 0.1 L = 0.05 moles

moles propanoic acid = molarity * volume

                                     = 0.75 M * 0.1 L = 0.075 moles

[NaOH] at equilibrium = 0.05 m

[propanoic acid ] at equilibrium = 0.075 - 0.05 = 0.025 m

when Pka for propanoic acid (given) = 4.89 

so by substitution:

∴PH = Pka + ㏒[NaOH]/[propanoic acid ]

∴ PH = 4.89 + ㏒ 0.05 / 0.025

         = 5.19
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Answer:

See explanation

Explanation:

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Percentage carbon = (7*12.01 amu/ 122.13 amu)*100% = 68.8%

Percentage oxygen = (2*16 amu/ 122.13 amu) *100% = 26.2%

c) Magnesium hydroxide (Mg(OH)2) has a molecular mass of 58.32 amu. It has 2 hydrogen atoms each with a mass of 1.01 amu; it has 1 magnesium atom with a mass of 24.3 amu and two oxygen atoms with a mass of 16.0 amu.

Percentage hydrogen = (2*1.01 amu/ 58.32 amu) *100% = 3.46 %

Percentage magnesium = (24.3 amu/58.32 amu)*100% = 41.7%

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Percentage nitrogen = (2*14.007 amu/ 60.064amu)*100% = 46.6%

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Percentage oxygen = (16 amu/60.064amu)*100% = 26.6%

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Percentage hydrogen = (14*1.01 amu/130.2 amu)*100% = 10.8%

Percentage carbon = (7*12.01 amu/130.2 amu)*100% = 64.6%

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Olegator [25]

A general exponential expression is something like:

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Using that we will get (-2)^6 = 64

With that definition, we can rewrite:

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So we just need to solve the above expression.

Also, remember the rule of signs:

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We will get:

(-2)*(-2)*(-2)*(-2)*(-2)*(-2) =  [(-2)*(-2)]*[(-2)*(-2)]*[(-2)*(-2)]

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