Part A Calculate the pH of 0.100 L of a buffer solution that is 0.23 M in HF (Ka

Question

Part A

Calculate the pH of 0.100 L of a buffer solution that is 0.23 M in HF (Ka = 3.5 x 10-4 ) and 0.45 M in NaF.

Express your answer using three significant figures.

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Part B

What is the pH after adding 0.001 mol of HNO3 to the buffer described in Part A?

Express your answer using three significant figures.

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Part C

What is the pH after adding 0.005 mol of KOH to the buffer described in Part A?

Express your answer using three significant figures.

Explanation / Answer

Part A

Calculate the pH of 0.100 L of a buffer solution that is 0.23 M in HF (Ka = 3.5 x 10-4 ) and 0.45 M in NaF.

Solution :-

Using the Henderson equation we can calculate the pH of the buffer

pH= pka + log [base/acid]

pka = -log ka

pka = -log 3.5*10^-4

pka = 3.46

pH= 3.46 + log [ 0.45 /0.23]

pH= 3.75

Part B

What is the pH after adding 0.001 mol of HNO3 to the buffer described in Part A?

Solution :-

Lets first calculate the moles of the NaF, HF and HNO3

Moles of NaF = 0.45 mol per L * 0.100 L = 0.045 mol

Moles of HF = 0.23 mol per L * 0.100 L = 0.023 mol

Moles of HNO3 = 0.001 mol

So after adding the HNO3 it will react with F- to form the HF

Therefore moles of F- after reaction = 0.045 – 0.001 = 0.044

Moles of HF after the reaction = 0.023 mol + 0.001 mol = 0.024 mol

New molarities

[F-] = 0.044 mol / 0.1 L = 0.44 M

[HF] = 0.024 mol / 0.100 L = 0.24 M

Now lets calculate the pH

pH= pka + log [base /acid]

pH= 3.46 + log [0.44/0.24]

pH= 3.72

Part C )

What is the pH after adding 0.005 mol of KOH to the buffer described in Part A?

Solution :-

Moles of HF = 0.023 mol

Moles of F- = 0.045 mol

After the reaction moles of F- = 0.045 mol +0.005 mol = 0.05 mol

Moles of HF = 0.023 mol – 0.005 mol = 0.018 mol

New molarities

[F-] = 0.05 mol / 0.100 L = 0.5 M

[HF] = 0.018 mol / 0.100 L = 0.18 M

Now lets calculate the pH

pH= pka + log [base /acid]

pH= 3.46 + log [0.5/0.18]

pH= 3.90

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