Answer:
THE pH OF THE SOLUTION IS 1.28
Explanation:
Step 1: Calculate the mass concentration of HCl
398 mg in 210 mL
1 L = 10^3 mL
1 g = 10^3 mg
398 *10^-3 g in 210 *10^-3 L
398 *10^-3 in 0.210 L
mass concentration = (0.398 * 1/ 0.210) g/dm3
= 1.895 g/dm3
Step 2: calculate the molar concentration
Molar mass of HCl = ( 1 + 35.5) = 36.5 g/mol
Molar concentration = mass concentration / RMM
Molar conc. = 1.895 / 36.5
Molar conc. = 0.0519 mol/dm3
So therefore, the molar concentartion of the solution is equal to the concentration of HCl.
In other words, [HCl] = 0.0519 mol/dm3
This shows that at 1 L of this solution contains 0.0519 moles of HCl
Step 3: write the dissiociation equation of HCl in water
HCl(aq) + H2O(l) --------> H30^+ (aq) + Cl- (aq)
Since HCl is a strong acid, it will dissociate completely in water forming the above products.
So we can say that:
[H3O+] = [HCl] = 0.0519 moles/ dm3
Step 4: calculate the pH
pH = -log[H3O+]
pH = -log (0.0519)
pH = 1.28(3 significant decimal place)
The pH of the solution is 1.28
