Chemistry: A 0.100 M solution of an acid, HA, has a pH = 2.00. What is the value of the ionization constant, Ka for this acid?

A 0.100 M solution of an acid, HA, has a pH = - id01002007349107, tchocho8621

PastelHibiscus

01.02.2022

$Ka=1.11x10^{-3}$

Explanation:

Hello,

In this case, since the ionization of the given HA acid is:

$HA\rightleftharpoons H^++A^-$

The equilibrium expression is:

$Ka=\frac{[H^+][A^-]}{[HA]}$

Whereas the concentration of hydrogen ions is compute from the pH=

$[H^+]=10^{-pH}=10^{-2.00}=0.01M$

Which also equals the concentration of A^- and the in general the ionization extent, therefore, the acid ionization constant, Ka, turns out:

$Ka=\frac{0.01*0.01}{0.1-0.01}\\ \\Ka=1.11x10^{-3}$

Regards.

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KariSupreme

24.04.2023

B. 3+

Explanation:

When Fe²⁺ loses an electron, it becomes an Fe³⁺ ion.

The loss of an electron in an atom makes the atom becomes positively charged. Further losing electrons makes the atom more positive.

For the given ion;

Fe²⁺ if it loses an electron more, it becomes Fe³⁺.

The expression is given as:

Fe²⁺ → Fe³⁺ + e⁻

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