Question

An aqueous solution is prepared by dissolving the following chemicals into distilled water: I) CaCl2 (8.0 mM); 3) HCl (2.0 mM). Please determine the pH of this solution at 25 °C Hint: Ionic strength effects are not negligible in this solution

Answer 1

Note: Ionic strength effects are not negligible for this solution

CaCl₂ dissolves in water as, CaCl₂ ----> Ca²⁺_(aq) + 2 Cl^-_(aq)

$\because$ 1.0 mM of CaCl₂ will produce 1.0 mM of Ca²⁺ ions and 2.0 mM of Cl^- ions.

$\therefore$ 8.0 mM of CaCl₂ will produce 8.0 mM of Ca²⁺ ions and (2.0x 8.0) = 16.0 mM of Cl^- ions.

CaCl₂ is an acidic salt (salt of strong acid + strong base), therefore will decrease pH of water i.e. will form H⁺ ions which will be equal in amount to the mM of Cl^- ions produced.

$\therefore$ 8.0 mM moles of CaCl₂ will produce 16.0 mM of H⁺ ions.

NaCl is a neutral salt (salt of strong acid + strong base), therefore will not alter the pH of water.

HCl will dissociate in water as HCl _(g) -------> H⁺_(aq) + Cl^-_(aq)

$\therefore$ 2.0 mM of HCl will produce 2.0 mM of H⁺ ions.

Now, the total mM of H⁺ ions produced on the addition of 8.0 mM CaCl₂, 4.0 mM NaCl and 2.0 mM HCl will be

Total mM of H⁺ = (Moles of H⁺ produced by CaCl₂ + Moles of H⁺ produced by HCl)

= 16.0 + 2.0 = 18.0 mM

$\therefore$ [H⁺] = 18.0 mM = 18.0 x 10^-3 M

$\because$ pH = - log₁₀ [H⁺]

$\therefore$ pH = - log₁₀ [18.0 x 10^-3]

= - [ log₁₀ 18.0 + log₁₀ 10^-3]

= - [ 1.255 + (-3 x log₁₀ 10) ]

= - [ 1.255 + (-3 x 1) ]

= - [ 1.255 - 3 ]

= - 1.255 + 3

= 3 - 1.255

$\therefore$ pH = 1.745

Answer: pH of solution at 25⁰C will be 1.745