Question

Calculate the pH of HCl and NAOH solutions after the addition of 0.75 L of water to: (a) 100 mL of 2.5 M HCl [2]

Answer 1

The pH of the new HCl solution is 0.53.

Step-by-step explanation:

Volume of the HCl solution
`V_1= 100 ml=0.1 L`

Molarity of HCl solution ,
`M_1= 2.5 M`

Volume of water added = 100 mL = 0.1 L

Molarity of HCl after addition of 0.75 L of water =
`M_2`

Final volume of the HCl solution after addition of 0.75 L of water =
`V_2`

`V_2=0.75 L + 0.1 L =0.85 L`

`M_1V_1=M_2V_2`[

`2.5 M* 0.1 L=M_2* 0.85 M`

`M_2=0.2941 M`

HCl is a monotonic acid and its one mole will give 1 mole of hydrogen ions in aqueous solution.

Molarity of hydrogen ion =
`[H^+]=0.2941 M`

`pH=-\log [H^+]=-\log [0.2941 M]=0.53`

The pH of the new HCl solution is 0.53.