Answer:
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Explanation:
Answer:
V2 = 35.967cm^3
Explanation:
Given data:
P1 = 0.2atm
P2 = 1.4atm
V1 = 250cm^3
V2 = ?
T1 = 10°C + 273 = 283K
T2 = 12°C + 273 = 285K
Apply combined law:
P1xV1/T1 = P2xV2/T2 ...eq1
Substituting values:
0.2 x 250/283 = 1.4 x V2/285
Solve for V2:
V2 = 14250/396.2
V2 = 35.967cm^3
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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