Question

23.1 g of HCl (a strong acid) is added to water to make 1250 mL of solution. Calculate [H3O+] and pH of the solution

Answer 1

Following are the solution to this question:

Step-by-step explanation:

`\to HCl \ mol = (23.1 \ g)/(36.5 (g)/(mol)) = 0.633 \ mole\\\\\\\to HCl \ molarity = ( 0.633 \ mole )/(1.25 L) \\\\`

`=0.5064 \ m`

`HCl +H_2O \longrightarrow H_3O^(+) + Cl^(-)\\\\\\`

`[H_3O^(+)] =[HCl] \\\\\to [H_3O^(+)] =0.5064 \ m \\\\\to PH =-\log [H_3O^(+)] \\\\`

`= -\log (0.5064)\\\\= 0.295 \\\\=0.3`