Question

Calculate the pH of the solution formed when 45.0 mL of 0.100M NaOH solution is added to 50.0 mL of 0.100M CH3COOH (Ka for acetic acid = 1.8 x10-5 ).

Answer 1

Answer:

pH of soltion will be 5.69

Explanation:

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH=pKa+log(\frac{[salt]}{[acid]} )$

$pKa=-log(Ka)$

$pKa=-log(1.8X10^{-5})=4.74$

The moles of acid taken :

$moles=molarityXvolume=0.1X50=5mmol$

The moles of base taken:

$moles=molarityXvolume=0.1X45=4.5mmol$

The moles of acid left after reaction :

$5-4.5=05mmol$

The moles of salt formed = 4.5mmol

Putting values in equation

$pH=pKa+log(\frac{[salt]}{[acid]} )=4.74+log(\frac{4.5}{0.5})=5.69$

Answer 2

The study of chemicals and bonds is called chemistry. There are two types of elements are there and these rare metals and nonmetals.

The correct answer is 5.59.

What is PH?

• pH, historically denoting "potential of hydrogen" is a scale used to specify the acidity or basicity of an aqueous solution.
• Acidic solutions are measured to have lower pH values than basic or alkaline solutions

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH= pka+log\frac{salt}{acid}$

$pka= -log(ka)$

$pka =-log(1.8*10^{-5}$

The moles of acid are taken as:-

$moles = M*V$

$0.5*50=5$

The moles of the base are taken as:-

$moles = M*V$

$0.1*45=4.5$

moles of acid left is 0.5

Place all the values to the equation:-

$pH=4.74+log\frac{4.5}{0.5} \\=5.69$

Hence, the correct answer is 5.69.

For more information about the pH, refer to the link:-

https://brainly.com/question/16759172