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Kingsley then adds 44.12 mL of NaOH to 250.00 mL of the HCOOH solution. The neutralization reaction resulted in 0.091 moles of HCOOH and 0.036 moles of HCOO- left in solution. Determine the pH of the resulting solution.

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Answer:

pH = 3.34

Explanation:

We can calculate the pH using Henderson-Hasselbach's equation:

  • pH = pKa + log[tex]\frac{[HCOO^-]}{[HCOOH]}[/tex]

for HCOOH, pKa = 3.75.

We can calculate [HCOO⁻] and [HCOOH] using the given number of moles for each one and the final volume:

  • Final Volume = 44.12 mL + 250 mL = 294 mL
  • 294 mL / 1000 = 0.294 L
  • [HCOO⁻] = 0.036 mol / 0.294 L = 0.122 M
  • [HCOOH] = 0.091 mol / 0.294 L = 0.310 M

Then we proceed to calculate the pH:

  • pH = 3.75 + log[tex]\frac{0.122}{0.310}[/tex]
  • pH = 3.34

The pH of the resulting solution is 3.34.

Calculation of the pH of the solution:

Since  

Final Volume = 44.12 mL + 250 mL = 294 mL

Now

= 294 mL / 1000

= 0.294 L

Now

[HCOO⁻] = 0.036 mol / 0.294 L = 0.122 M

[HCOOH] = 0.091 mol / 0.294 L = 0.310 M

So, the pH should be

= 3.75 + log0.122/0.310

= 3.34

hence, The pH of the resulting solution is 3.34.

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