2.643 grams of potassium butanoate (KCH3(CH2)2CO2 ) is fully dissolved in 50.00 mL of water, which is carefully transferred to a conical flask. Then 100.00 mL of 0.120 M HCℓ is added dropwise to this solution from a burette. Given: Ka(butanoic acid) = 1.5 × 1O−5 . 2.1 Showing all your calculations and reasoning, determine the pH of the solution that results after the addition of all the acid mentioned above.

The pH of the solution is 4.69

Explanation:

Given that,

Mass of potassium = 2.643 grams

Weight of water = 50.00 mL

Weight of HCl=100.00 ml

Mole = 0.120 M

We know that,

is a basic salt.

Let's write it as KY.

The acid would become HY.

We need to calculate the moles of KY

Using formula of moles

The reaction is

The number of moles of KY is 20.98 m

initial moles = 20.98

Final moles

We need to calculate the value of pKa(HY)

Using formula for pKa(HY)

We need to calculate the pH of the solution

Using formula of pH

Put the value into the formula

Hence, The pH of the solution is 4.69

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