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

Answer:

Following are the solution to this question:

Explanation:

$\to HCl \ mol = \frac{23.1 \ g}{36.5 \frac{g}{mol}} = 0.633 \ mole\\\\\\\to HCl \ molarity = \frac{ 0.633 \ mole }{1.25 L}$

                            $=0.5064 \ m$

 $HCl +H_2O \longrightarrow H_3O^{+} + Cl^{-}$

$[H_3O^{+}] =[HCl] \\\\\to [H_3O^{+}] =0.5064 \ m \\\\\to PH =-\log [H_3O^{+}]$

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