The equilibrium constant of reaction, usually denoted as K, is a unit of ratio. The ratio involves concentrations of products to reactants. But you also have to incorporate their stoichiometric coefficients in the reaction as their respective exponents. Note that substances in their aqueous state are the ones that are included only in the expression. To properly show you how it's done, consider this equilibrium reaction:
aA (aq) + bB (l) ⇆ nN (aq)
Since only reactant A and product N are aqueous, the equilibrium constant for this reaction is:
K = [N]ⁿ/[A]ᵃ
where the [] brackets denotes concentration in molarity
Now, let's apply this to the given equation:
Cr₂O²⁻ (aq) + 6 I⁻ (aq) + 14 H⁺ (aq) → 3 I₂ (s) + 2 Cr³⁺ (aq) + 7 H₂O<span> (l)
</span>I think there is a typographical error because Cr₂O²⁻ has a negative 2 charge rather than -27. Remember that only substances in aqueous states are included in the K expression. Therefore, the expression for K is:
K = [Cr³⁺]² / [Cr₂O²⁻][I⁻]⁶[H⁺]¹⁴
Answer:
0.375 L
Explanation:
We know that at neutralization, the number of mol of acid must equal the number of equivalents of base.
This is a reaction 1:1 acid to base:
HClO₄ + NaOH ⇒ NaClO₄ + H₂O
We re given the moles of the base indirectly since we know the volume and molarity. From there we can calculate the volume of HClO₄.
Moles NaOH = 0.115 L x 0.244 M = 0.115 L x 0.244 mol/L =0.028 mol
Thus we require 0.028 mol of HClO₄ in the pechloric acid solution:
Molarity = # moles / V ⇒ V = # moles / M
V = 0.028 mol / 0.0748 mol/L = 0.375 L
Note that this problem can be solved in just one step since
M(HClO₄) x V(HClO₄) = M(NaOH) x V(NaOH) ⇒
V(HClO₄) = M(NaOH) x V(NaOH) / M(HClO₄)
Answer:
pH = 6,951
Explanation:
The neutral form of aspartic acid is in its isolectric point. For aspartic acid the isoelectric point is the average of pka1 and pka2, thus:
= 2,945
The addition of stoichiometric amounts of NaOH at the first equivalence point will increase the pH at average of pka2 and pka3 thus:
= 6,951
Thus, <em>pH at the first equivalence point is 6,951</em>
I hope it helps!
Mass of sodium is 23 and mass of fluirine is 19 there mass of NaF is 42
42 g = 1 mole therefor 4.5 moles will have
4.5 × 42 = 189 g
