Question

A chemist dissolves 398. Mg of pure hydrochloric acid in enough water to make up 210. ML of solution. Calculate the pH of the solution. Round your answer to 3 significant decimal places.

Answer 1

THE pH OF THE SOLUTION IS 1.28

Step-by-step 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