Answer:
5.2 x 10⁻⁴ M.
Explanation:
- The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:
<em>P = kC</em>
where P is the partial pressure of the gaseous solute above the solution.
k is a constant (Henry’s constant).
C is the concentration of the dissolved gas.
- At two different pressures, there is two different concentrations of dissolved gases and is expressed in a relation as:
<em>P₁C₂ = P₂C₁,</em>
P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.
P₂ = 0.76 atm, C₂ = ??? mol/L.
<em>∴ C₂ = (P₂C₁)/P₁ =</em> (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = <em>5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.</em>
Answer:
11.39
Explanation:
Given that:
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
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