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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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Answer:

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HCl is a strong acid that dissociates according to the following equation.

HCl ⇒ H⁺ + Cl⁻

The molar ratio of HCl to H⁺ is 1:1. The concentration of H⁺ is 1/1 × 0.00013 M = 0.00013 M.

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[OH⁻] = 10⁻¹⁴/[H⁺] = 10⁻¹⁴/0.00013 = 7.7 × 10⁻¹¹ M

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Answer:

1.4 × 10^-4.

Explanation:

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So, from the question above we are given the following parameters or data which is going to help in solving this particular Question/problem;

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