Question

what will be the ph of the solution if 0.01 moles of hcl dissolved in a buffer solution containing 0.02 mole of propanoic acid(ka=1.34×10^-5) and 0.0152moles of salt,at 25°C
(log0.173=-0.76)

Answer 1

Answer:- 4.11

Solution:- The buffer solution has weak propanoic acid and it's salt that is it's conjugate base. When an acid like HCl is added to the buffer then it reacts with the conjugate base of the acid(propanoic acid). In this reaction the moles of conjugate base decreases and the weak acid moles increases. The reaction is shown below:

$CH_3CH_2COO^-+H^+\leftrightharpoons CH_3CH_2COOH+H_2O$

The reaction is taking place in 1:1 mol ratio. From given information, 0.01 mol of HCl are added to the buffer that has 0.02 moles of propanoic acid and 0.0152 moles of it's conjugate base.

So, moles of conjugate base after addition of HCl = 0.0152 - 0.01 = 0.0052

and moles of propanoic acid after addition of HCl = 0.02 + 0.01 = 0.03

We use Handerson equation to calculate the pH of the buffer solutions. The equation is:

$pH=pKa+log(\frac{base}{acid})$

let's first calculate the pKa from given Ka value using the equation:

$pKa=-logKa$

$pKa=-log1.34*10^-^5$

pKa = 4.87

Let's plug in the values in Handerson equation now to calculate the pH:

$pH=4.87+log(\frac{0.0052}{0.03})$

pH = 4.87 + log(0.173)

pH = 4.87 - 0.76

pH = 4.11

So, the pH of the buffer solution after an addition of 0.01 moles of HCl is 4.11.

Answer 2

Answer:

Solution:- The buffer solution has weak propanoic acid and it's salt that is it's conjugate base. When an acid like HCl is added to the buffer then it reacts with the conjugate base of the acid(propanoic acid). In this reaction the moles of conjugate base decreases and the weak acid moles increases. The reaction is shown below:

CH_3CH_2COO^-+H^+\leftrightharpoons CH_3CH_2COOH+H_2OCH

3

CH

2

COO

−

+H

+

⇋CH

3

CH

2

COOH+H

2

O

The reaction is taking place in 1:1 mol ratio. From given information, 0.01 mol of HCl are added to the buffer that has 0.02 moles of propanoic acid and 0.0152 moles of it's conjugate base.

So, moles of conjugate base after addition of HCl = 0.0152 - 0.01 = 0.0052

and moles of propanoic acid after addition of HCl = 0.02 + 0.01 = 0.03

We use Handerson equation to calculate the pH of the buffer solutions. The equation is:

pH=pKa+log(\frac{base}{acid})pH=pKa+log(

acid

base

)

let's first calculate the pKa from given Ka value using the equation:

pKa=-logKapKa=−logKa

pKa=-log1.34*10^-^5

pKa = 4.87

Let's plug in the values in Handerson equation now to calculate the pH:

pH=4.87+log(\frac{0.0052}{0.03})pH=4.87+log(

0.03

0.0052

)

pH = 4.87 + log(0.173)

pH = 4.87 - 0.76

pH = 4.11

So, the pH of the buffer solution after an addition of 0.01 moles of HCl is 4.11.