Question

Determine the ionic strength, ???? , for each of the solutions. Assume complete dissociation of each salt and ignore any hydrolysis reactions. A solution of 0.00508 M HCl . ????= 5.08 ×10 −3 M A solution of 0.00127 M LaBr3 . ????= 7.62 ×10 −3 M A solution of 0.000870 M CaBr2 and 0.000784 M NaBr .

Answer 1

HCl I = 0.00508 M

LaBr₃ I = 0.00762 M

CaBr₂ + NaBr I = 0.00339 M

Step-by-step explanation:

Ionic strength (I) is a measure of the concentration of ions in a solution. It can be calculated with the following expression:

I = 1/2 ∑Ci × Zi²

where,

Ci is the molarity of each ion (molarity of the salt × amount of each ion in the molecular formula)

Zi is the charge of each ion

A solution of 0.00508 M HCl

HCl ⇒ H⁺ + Cl⁻

I = 1/2 × [C(H⁺) × Z(H⁺)² + C(Cl⁻) × Z(Cl⁻)²]

I = 1/2 × [0.00508M × 1² + 0.00508M × (-1)²]

I = 0.00508 M

A solution of 0.00127 M LaBr₃

LaBr₃ ⇒ La³⁺ + 3 Br⁻

I = 1/2 × [C(La³⁺) × Z(La³⁺)² + C(Br⁻) × Z(Br⁻)²]

I = 1/2 × [0.00127 M × 3² + 3 × 0.00127 M × (-1)²]

I = 0.00762 M

A solution of 0.000870 M CaBr₂ and 0.000784 M NaBr

Here we have 2 sources of ions with a common ion that is Br⁻. To simplify, we will compute every ion from each salt separately, in the order they are presented.

CaBr₂ ⇒ Ca²⁺ + 2 Br⁻

NaBr ⇒ Na⁺ + Br⁻

I = 1/2 × [C(Ca²⁺) × Z(Ca²⁺)² + C(Br⁻)₁ × Z(Br⁻)₁²+C(Na⁺) × Z(Na⁺)² + C(Br⁻)₂ × Z(Br⁻)₂²]

I = 1/2 × [0.000870 M × 2² + 2 × 0.000870 M × (-1)² + 0.000784 M × 1² + 0.000784 M × (-1)²]

I = 0.00339 M