formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer

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24-07-2020
Chemistry

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formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer

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jufsanabriasa jufsanabriasa · Answer 1 · 2020-07-25T03:14:20+02:00

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

tallinn tallinn · Answer 2 · 2022-04-07T14:21:29+02:00

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:

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