Part A a solution that is 0.160 M in propanoic acid and 0.125 M in potassium pro

Question

Part A

a solution that is 0.160 M in propanoic acid and 0.125 M in potassium propanoate

Express your answer using two decimal places.

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Part B

a solution that contains 0.800% C5H5N by mass and 0.880% C5H5NHCl by mass

Express your answer using two decimal places.

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Part C

a solution that is 14.5 g of HF and 26.5 g of NaF in 125 mL of solution

Express your answer using two decimal places.

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pH =

Explanation / Answer

Solution

Part A)0.160 M propanoic acid and 0.125 M sodium propanote

Using the Henderson equation we can calculate the pH

pH= pka + log ([base]/[acid])

pka of the propanoic acid = 4.89

pH= 4.89 + log[0.125/0.160]

pH= 4.89 + 0.107

pH= 5.00

Part B)

0.800% C5H5N by mass and 0.880% C5H5NHCl

Assume that we have 0.800 g C5H5N and 0.880 g C5H5NHCl

Now lets calculate moles of each

Moles = mass / molar mass

Moles of C5H5N = 79.1 g per mol = 0.0101 mol

Moles of C5H5NHCl = 0.880 g / 115.56 g per mol = 0.00762 mol

Now using the Henderson equation we can calculate its pH

pKb of pyridine =8.77

Now lets calculate the pOH and then pH

pOH= pkb + log ([acid]/[base])

pOH = 8.77 + log[0.00762/0.0101]

pOH =8.77+(-0.122)

pOH = 8.65

part C) 14.5 g HF and 26.5 NaF

lets calculate the moles of each

moles of HF = 14.5 g / 20.0 g per mol = 0.725 mol HF

moles of NaF = 26.5 g / 41.988 g per mol = 0.631 mol NaF

now lets calculate the pH using the Henderson equation

pH= pka + log ([base]/[acid])

pka of HF = 3.14

pH = 3.14 + log [0.631/0.725]

pH= 3.14+ (-0.06)

pH= 3.08

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