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jekas [21]
4 years ago
12

Question 14 (5 points)

Chemistry
1 answer:
Svetach [21]4 years ago
5 0

Answer:

D) 2.01 x 10⁻⁴ .

Explanation:

pH = 2.11

[ H⁺ ] = 10^{-2.11}  

Let the acid be HA

It will ionise as follows .

                                        HA       ⇄       H⁺       +        A⁻

in equilibrium                 .30               10^{-2.11}          10^{-2.11}         

Acid ionisation constant Ka  =   \frac{(10^{-2.11})^2}{0.3}

= 2 x 10⁻⁴                

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