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

The pH of the solution is 4.69

Step-by-step 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=(m)/(M)*1000`

`moles\ of\ KY=(2.643)/(126)*1000`

`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*100= 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*10^(-5))`

`pKa_(HY)=4.82`

We need to calculate the pH of the solution

Using formula of pH

`pH=pKa+log(([KY])/([KH]))`

Put the value into the formula

`pH=4.82+log((8.98)/(12))`

`pH=4,69`

Hence, The pH of the solution is 4.69