Question
Asked Dec 5, 2019
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A solution is made by mixing exactly 500 mL of 0.156 M NaOH with exactly 500 mL of 0.100 MCH3COOH. Calculate the equilibrium concentration of the species below.

Ka of CH3COOH is 1.8
×
10−5


 

[
 
H+
]


 

 
×

10

 

M
Enter your answer in scientific notation.
 
 

[
 
OH
]


 

 

M
 
 

[
 
CH3COOH
]


 

 
×

10

 

M
Enter your answer in scientific notation.
 
 

[
 
Na+
]


 

 

M
 
 

[
 
CH3COO
]


 

 

M

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Expert Answer

Step 1

The relation between milliliters (mL) and liters (L) is written below.

1 mL = 0.001 L

Convert  500 mL into liters by using the above relation.

500 mL = 0.500 L

 The number of moles of NaOH is calculated in equation (1).

      …… (1)

 

The number of moles of CH3COOH  is calculated in equation (2).

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M (1) 0.156 M = 0.500 L n= 0.078 mole M = п (2) 0.100 M 0.500 L n=0.05 mole

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Step 2

The number of moles of NaOH left in the solution is calculated below.

n = 0.078 mol – 0.05 mol

   = 0.028 mol

 

The total volume of the solution is 1000 mL or 1 L. In 1 L solution, the concentration of species becomes equal to the number of moles. The total concentration of sodium ions in the solution is from sodium hydroxide(0.028 M) and sodium acetate (0.05 M). Thus,  the concentration of sodium ion is 0.078 M.

 

NaOH is a strong electrolyte as it completely ionizes itself. The left over concentration of  hydroxide ion is  0.028 M.

Step 3

 The hydrogen concentration ...

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Image Transcriptionclose

[н Тон ]-10ч [н] =1014 1014 H* Гон [н] ОН 10-14 Го.028] — 3.6х10-13 М

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