PbI(ii) ionization in the solution of PBI(ii) into water is:
<span>PbI</span>₂(solution) <==> Pb₂⁺ + 2I⁻
If the conc. of PbI(ii) in the sol. is xM then the conc. of Lead(ii) will be x M and conc. of iodide will be 2 x M.
Therefore,
<span>Ksp=<span>[Pb</span></span>²⁺][I-]²
Plugging the values:
1.4×10⁻⁸ = x ⋅ (2x)²
1.4×10⁻⁸ = 4x³
x³ = {1.4×10⁻⁸}÷4
x³ = 0.35 x 10⁻⁸
or
x³ = 3.5 x 10⁻⁹
x = 1.51 x 10⁻³
Hence,
Concentration of iodide ions in the solution:
2x = 3.02 x 10⁻³
<h3>
Answer: A) 3.5 mol/L</h3>
Explanation:
To determine the molarity, we have to find the number of moles in the volume given, and then extrapolate to find the number of moles that would be in 1 L.
<u>Determine the moles in the given volume</u>
moles of LiCl = mass ÷ molar mass
= 139.9 g ÷ 42.39 g/mol
= 3.30 mol
<u>Find the moles in 1 L</u>
Since 930 mL of LiCl = 3.30 mol
then 1000 mL of LiCl = (3.30 mol × 1000 mL/L) ÷ 930 mL
= 3.55 mol/L
Answer:
Explanation:
Sr(OH)2 (aq) ⇔ Sr+2 (aq) + 2OH- (aq)
Answer:
Phenol red indicator.
Explanation:
- Phenol red indicator is an organic dye which works as an indicator in the pH range (6.8 - 8.4) for acid-base reactions.
<em>The color in acidic solutions is yellow and in basic solution is red.</em>
Ionic compounds form between metals and non metals.
