Question

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.

Answer 1

Answer:

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,

$KCH_{3}(CH_{2})_{2}CO_{2}$ is a basic salt.

Let's write it as KY.

The acid $HCH_{3}(CH_{2}CO_{2})$ would become HY.

We need to calculate the moles of KY

Using formula of moles

$moles\ of\ KY=\dfrac{m}{M}\times1000$

$moles\ of\ KY=\dfrac{2.643}{126}\times1000$

$moles\ of\ KY=20.97\ m\ mole$

The reaction is

$KY+HCl\Rightarrow HY+ KCl$

The number of moles of KY is 20.98 m

initial moles = 20.98

Final moles $m=20.98-0.120\times100= 8.98$

We need to calculate the value of pKa(HY)

Using formula for pKa(HY)

$pKa_{HY}=-log Ka$

$pKa_{HY}=-log(1.5\times10^{-5})$

$pKa_{HY}=4.82$

We need to calculate the pH of the solution

Using formula of pH

$pH=pKa+log(\dfrac{[KY]}{[KH]})$

Put the value into the formula

$pH=4.82+log(\dfrac{8.98}{12})$

$pH=4,69$

Hence, The pH of the solution is 4.69