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What is the pH of a 4.5 x 10^-3 M HI solution?

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Answer: 2.4

Step-by-step explanation:

User Avo
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To find the pH of a 4.5 x 10^-3 M HI solution, we need to know the dissociation constant of HI (also known as the acid ionization constant, Ka). The Ka of HI is 1.3 x 10^-10.

HI ⇌ H+ + I-

At equilibrium, let x be the concentration of H+ ions, then [H+] = x and [I-] = x. The initial concentration of HI is 4.5 x 10^-3 M. Using the Ka expression, we can set up the following equation:

Ka = [H+][I-]/[HI]

1.3 x 10^-10 = (x)(x)/(4.5 x 10^-3)

Solving for x, we get:

x = 7.07 x 10^-6 M

Since pH = -log[H+], we can calculate the pH of the solution:

pH = -log(7.07 x 10^-6) = 5.15

Therefore, the pH of a 4.5 x 10^-3 M HI solution is 5.15.

User Robert Robinson
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