Respuesta :
Answer:
pH = 3.95
Explanation:
It is possible to calculate the pH of a buffer using H-H equation.
pH = pka + log₁₀ [HCOONa] / [HCOOH]
If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:
3.77 = pka + log₁₀ [0.50] / [0.50]
3.77 = pka
Knowing pKa, the NaOH reacts with HCOOH, thus:
HCOOH + NaOH → HCOONa + H₂O
That means the NaOH you add reacts with HCOOH producing more HCOONa.
Initial moles of 100.0mL = 0.1000L:
[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH
[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa
After the reaction, moles of each species is:
0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH
0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa
With these moles of the buffer, you can calculate pH:
pH = 3.77 + log₁₀ [0.0600] / [0.0400]
pH = 3.95
When the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer pH is = 3.77 + log₁₀ [0.0600] / [0.0400] = 3.95
What is Formic Acid?
It is possible to Computation the pH of a buffer using H-H equation.
Then pH is = pka + log₁₀ [HCOONa] / [HCOOH]
Then If concentration of [HCOONa] is = [HCOOH] then = 0.50M and pH = 3.77:
3.77 is = pka + log₁₀ [0.50] / [0.50]
After that, 3.77 = pka
Then, Knowing pKa, the NaOH reacts with HCOOH, thus:
After that,[tex]HCOOH + NaOH \rightarrow HCOONa + H2O[/tex]
Now, That means the NaOH you add reacts with HCOOH producing more HCOONa.
Then, Initial moles of 100.0mL = 0.1000L:
After that, [HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH
Then, [HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa
After that, when the reaction, moles of each species is:
Then, 0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH
Now, 0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa
Then, With these moles of the buffer, you can calculate pH:
pH = 3.77 + log₁₀ [0.0600] / [0.0400]
Therefore, pH = 3.95
Find more information about Formic Acid here:
https://brainly.com/question/26708431
