Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:
[tex]HCOOH\leftrightarrow H^++HCOO^-[/tex]
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:
[tex]\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]
[tex]1.8*10^-^4=\frac{x^2}{c-x}[/tex]
From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,
[tex]1.8*10^-^4=\frac{x^2}{0.12}[/tex]
Let's solve this for x. Multiply both sides by 0.12
[tex]2.16*10^-^5=x^2[/tex]
taking square root to both sides:
[tex]x=0.00465[/tex]
Now, we have got the concentration of [tex][H^+] .[/tex]
[tex][H^+] = 0.00465 M[/tex]
We know that, [tex]pH=-log[H^+][/tex]
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
Answer:
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:
[tex]K_{a} = \frac{[H^{+} ][HCOO^{-} ]}{[HCOOH]}[/tex]
[tex]1.8 x 10^{-4} = \frac{(x x)}{(0.12 M -x)}[/tex]
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= [tex]\sqrt{0.12 x 1.8 x 10^{-4} }[/tex]= 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
