Answer:
The concentration of KOH is 0.186 M
Explanation:
First things first, we need too write out the balanced equation between HBr and KOH.
This is given as;
KOH (aq) + HBr (aq) → KBr (aq) + H2O (l)
From the reaction above, we can tell that it takes 1 mole of KOH to react with 1 mole of HBr.
We use the acid base formular in calculating unknown concentrations. This is given as;
where;
Ca = Concentration of acid
Va = Volume of acid
Cb = Concentration of base
Vb = Volume of base
na = Number of moles of acid
nb = Number of moles of base
KOH is the base and HBr is acid.
Hence;
Ca = 0.225
Va = 35
Cb = ?
Vb = 42.3
na = 1
nb = 1
Making Cb subject of formular we have;
Cb = (0.225 * 35 * 1) / (42.3 * 1)
Cb = 0.186 M
Mass of Co(NO₃)₂ = 1.95 g
V KOH = 0.350 L
[KOH] = 0.220 M
Kf = 5.0 x 10⁹
molar mass of Co(NO₃)₂ = 182.943 g/mol
so [Co(NO₃)₂] = 1.95 / (0.350 * 182.943) = 0.03045 M
[Co²⁺] = 0.03045 M
[OH⁻] = 0.22 M
chemical reaction:
Co²⁺(aq) + 4 OH⁻ ⇄ Co(OH)₄²⁻
I (M) 0.03045 0.22 0
C (M) - 0.03045 - 4 (0.03045) 0.03045
E (M) - x 0.22 - 4(0.03045) 0.03045
= 0.0982
Kf = [Co(OH)₄²⁻] / [Co⁺²][OH⁻]⁴
5.0 x 10⁹ = (0.03045) / x (0.0982)⁴
x = 6.5489 x 10⁻⁸
at equilibrium:
[Co²⁺] = 6.54 x 10⁻⁸
[OH⁻] = 0.0982 M
[Co(OH)₄²⁻] = 0.03045 M
Covalent Bond.
To be specific, it is polar covalent bond. :)