Question

calculate the pH of 100 ml M /2 HCL +100 ml M/4 Naoh​

Answer 1

Given: Volume of HCL = 100ml

Molarity of HCL = M/2

Volume of NaoH = 100ml

Molarity of NaoH = M/4

To Find :- pH of solution of HCL and NaoH

Solution:-

- As we know,

Molarity = no. of moles / volume in Ltr

So, no. of moles = M × V (in L )

- Given, Molarity of HCl = M/2 means 1/2 moles of solute in 1 L of solution.

So, M (HCl) = 1/2 = 0.5 M

and, Volume of HCl = 100 ml = 0.1 L

- Thus, no. of moles of HCl

= 0.1 × 0.5

= 0.05 moles

- Similarly, Molarity of NaoH = M/4 means 1/4 moles of solute in 1 L of solution.

So, M(NaoH) = 1/4 = 0.25 M

and, Volume of NaoH = 100 ml = 0.1 L

- Thus, no. of moles of NaoH

= 0.1 ×0.25

= 0.025 moles

- The neutralization reaction is,

NaoH + HCl ---> NaCl + H2O

- To neutralize 0.025 mole of NaCl the amount of HCl required is

= 0.025

- The HCl left after neutralization as it present in excess is,

= 0.05 - 0.025

= 0.025 moles

- Total volume of the HCl solution is 1L

- So, the Concentration of solution

= no. of moles left / Volume of solution in L

= 0.025/1

= 0.025 M or 25 × 10 ^(-3) M

- As we know that,

pH- log [H^+]

pH = -log [25 × 10 ^(-3) ]

pH = -[ 1.3979 -3]

pH = -[-1.6021]

pH = 1.6021

- Hence, The pH of the solution is 1.6021.