Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

Let's solve this for x. Multiply both sides by 0.12

taking square root to both sides:

Now, we have got the concentration of

We know that,

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

The correct answer is 2.34

Explanation:

HCOOH is formic acid. It is a weak acid so it does not dissociates completely in water. At the beggining (I) the initial concentration is 0.12 M. In water it will dissociate in a certain grade x as follows:

          HCOOH    →    H⁺ + HCOO⁻

I             0.12 M           0          0                    

C           - x                   x          x

E          (0.12 M - x)       x          x

The mathematical expression for the equilibrium constant (Ka) is the following:

As the value of Ka is too small in comparison with the initial concentration 0.12 M, we can approximate: 0.12 M - X ≅ 0.12 M. Then, we calculate x:

1.8 x 10⁻⁴ = x²/0.12 M

⇒ x= = 4.65 x 10⁻³

Since x = 4.65 x 10⁻³ , from the equilibrium we have:

[H⁺] = x = 4.65 x 10⁻³

From the definition of pH, we have:

pH = -log [H⁺] = -log (4.65 x 10⁻³)= 2.34

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

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