A.) How many moles of sodium hydroxide would have to be added to 150 mL of a 0.4

Question

A.) How many moles of sodium hydroxide would have to be added to 150 mL of a 0.479 M nitrous acid solution, in order to prepare a buffer with a pH of 3.340?

moles

B.) An aqueous solution contains 0.427 M hydrofluoric acid.

How many mL of 0.250 M sodium hydroxide would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 3.250?

mL

C.) An aqueous solution contains 0.437 M ammonia (NH3).

How many mL of 0.397 M hydrobromic acid would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 8.870?

mL

Explanation / Answer

A.) How many moles of sodium hydroxide would have to be added to 150 mL of a 0.479 M nitrous acid solution, in order to prepare a buffer with a pH of 3.340?

pka HNO2 = 3.39.

The formula for pH of a buffer is:

pH = pKa + log (salt/acid)

Since the volume of the solution will be the same in the solution we can use moles in the equation:

pH = pKa + log (moles salt/moles acid)

The reaction that takes place is:

HNO2 + NaOH -> NaNO2 + H2O

pH = pH = pKa + log (moles NaNO2 / moles HNO2)

Then, the moles we add of NaOH will become moles of NaNO2 as well as moles of HNO2 at the end will be the initial moles of HNO2 minus the moles of NaOH:

moles HNO2 = initial moles HNO2 - moles NaOH

moles NaNO2 = moles NaOH

Then the equation changes to:

pH = pH = pKa + log (moles NaOH / initial moles HNO2 - moles NaOH)

The initial moles of HNO2 are: 0.15L(0.479mol/L) = 0.07185.

So, the equation becomes:

pH = pKa + log (moles NaOH / 0.07185 - moles NaOH)

Since pKa = 3.39 and pH = 3.34:

3.39 = 3.34 + log(moles NaOH / 0.07185 - moles NaOH)

0.05 = log (moles NaOH / 0.07185 - moles NaOH)

1.12 = (moles NaOH / 0.07185 - moles NaOH)

1.12 (0.07185 - moles NaOH) = moles NaOH

0.0806 - 1.12 moles NaOH = moles NaOH

0.0806 = 2.12 moles NaOH

moles NaOH = 0.0380 moles NaOH

B.) An aqueous solution contains 0.427 M hydrofluoric acid.
How many mL of 0.250 M sodium hydroxide would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 3.250?

pKa HF = 3.14

The formula for pH of a buffer is:

pH = pKa + log (salt/acid)

Since the volume of the solution will be the same in the solution we can use moles in the equation:

pH = pKa + log (moles salt/moles acid)

The reaction that takes place is:

HF + NaOH -> NaF + H2O

pH = pH = pKa + log (moles NaF / moles HF)

Then, the moles we add of NaOH will become moles of NaF as well as moles of HF at the end will be the initial moles of HF minus the moles of NaOH:

moles HF = initial moles HF - moles NaOH

moles NaF = moles NaOH

Then the equation changes to:

pH = pH = pKa + log (moles NaOH / initial moles HF- moles NaOH)

The initial moles of HF are: 0.15L(0.25mol/L) = 0.0375

So, the equation becomes:

pH = pKa + log (moles NaOH / 0.0375 - moles NaOH)

Since pKa = 3.14 and pH = 3.25:

3.25 = 3.14 + log(moles NaOH / 0.0375 - moles NaOH)

0.11 = log (moles NaOH / 0.0375 - moles NaOH)

1.29 = (moles NaOH / 0.0375 - moles NaOH)

1.29 (0.0375- moles NaOH) = moles NaOH

0.0483 - 1.29 moles NaOH = moles NaOH

0.0483 = 2.29 moles NaOH

moles NaOH = 0.0211 moles NaOH

Note: we're only allowed to answer 1 question per Q&A. Since A) and B) had very similar procedures I did both. Please ask for C) in another Q&A.

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