The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x
Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M
Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x
We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4
We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10
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Explanation:
Answer:
Δ
N = -1
Explanation:
Step 1: Data given
2 SO2(g) + O2(g) ⇌ 2 SO3(g)
For 2 moles of SO2 we need 1 mol of O2 to produce 2 moles of SO3
Step 2: Calculate Δ
Δ
N wNill be the sum of moles of products minus the sum of moles of reactants.
Number of moles of products = number of moles of SO3 = 2
Number of moles of reactants = number of moles of SO2 + number of moles of O2 = 2 + 1 = 3
Δ
N = 2 - 3 = -1
