Question

3 gram of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken is another container 1/2 mL of 5M NaOH is added to this. Find the pH of this solution. Find the pH of this solution. (log 3 = 0.4771, pKa = 4.74)​

Answer 1

3 gram of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken is another container 1/2 mL of 5M NaOH is added to this. Ph is calculated as given below

1. m mole of acidic acid in 20 mL = 2

m mole of HCl in 20 mL = 1

m mole of NaOH = 2.5

2. pH = PKa + log(3/2)/2

= 4.74 + log 3

= 4.74 + 0.48 = 5.22

Answer 2

The ph of the solution can be calculated as follows:

We know that :

$molarity =\frac{moles}{Vol(L)}$

Using the above formula,

milli moles of acetic acid in 20 ml = 2

milli moles of HCL in 20 ml = 1

milli moles of NaOH = 2.5

From Henderson's equation :

$ph= pka + log\frac{\frac{3}{2} }{2}$

ph = 4.74 + log 3

     =4.74 + 0.48

     = 5.22