A chemist dissolves 327.mg of pure hydrochloric acid in enough water to make up 120.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant dig

pH → 1.13

Explanation:

Our solution is pure HCl

HCl(aq) + H₂O(l)  →  H₃O⁺(aq) + Cl⁻(aq)

As a strong acid, it is completely dissociated.

1 mol of HCl, can give 1 mol of H⁺ to the medium. Water does not participate. Let's find out M for the acid.

1st step: We convert the mass from mg to g → 327 mg . 1g /1000mg = 0.327 g

2nd step: We convert the mass(g) to moles: 0.327 g / 36.45 g/mol = 8.97×10⁻³ moles

3rd step: We convert the volume from mL to L → 120mL . 1L /1000 mL = 0.120L

Molarity (mol/L) = 8.97×10⁻³ mol / 0.120L = 0.075M

We propose: HCl(aq) + H₂O(l)  →  H₃O⁺(aq) + Cl⁻(aq)

                     0.075M                    0.075M

pH = - log [H₃O⁺] → - log 0.075 = 1.13 → pH

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