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Svetach [21]
3 years ago
5

If the ka of a monoprotic weak acid is 8.4 × 10-6, what is the ph of a 0.45 m solution of this acid

Chemistry
1 answer:
olga_2 [115]3 years ago
7 0

Consider the acid to be HA so the equation for acid will be:

HA+H₂O---->H₃O⁺ +A⁻

Initial: 0.45 0 0 0

Final 0.45-x x x x

Kₐ= [H₃O⁺][A⁻]/[HA]

= x²/0.45-x

8.4 × 10⁻⁶= x²/0.45-x

3.78× 10⁻⁶-8.4 × 10⁻⁶x=x²

On solving equation,

x= 0.0018

[H₃O⁺]=[A⁻]=x=0.0018

pH= -log[H₃O⁺]

= -log[0.0018]

= 2.74

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A chemical reaction is shown below:
BabaBlast [244]

Answer:

Mass = 8.46 g

Explanation:

Given data:

Mass of water produced = ?

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Mass of oxygen = 15 g

Solution:

Chemical equation:

C₆H₁₂O₆ + 6O₂     →   6H₂O + 6CO₂

Number of moles of glucose:

Number of moles = mass/molar mass

Number of moles = 20 g/ 180.16 g/mol

Number of moles = 0.11 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 15 g/ 32 g/mol

Number of moles = 0.47 mol

now we will compare the moles of water with oxygen and glucose.

               C₆H₁₂O₆           :            H₂O

                   1                   :              6

                 0.11                :           6/1×0.11 = 0.66

                   O₂               :            H₂O

                   6                   :              6

                 0.47                :           0.47

Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 0.47 mol  ×18 g/mol

Mass = 8.46 g

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3 years ago
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Answer:

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Answer:

a

Explanation:

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