Respuesta :

missyvee
We are given a pH of 10.38
*pH + pOH = 14

10.38 + pOH = 14
----> pOH = 3.62

pOH = -log [OH]
----> [OH] = 2.3988x10^-4

Set up an I.C.E table  using the concentration of OH found (y) and x being the unknown:
      [base] + H2O ------> [conjugate acid] + OH
I         x                                      0                      0
C      -y                                     +y                     +y
E       x-y                                    y                       y

Given Kb, we can set up the equilibrium equation:
K = [tex] \frac{prdoucts}{reactants} [/tex]
K = [tex] \frac{y ^{2} }{x-y} [/tex]

We can substitute y for the concentration of OH we found, and K for the Kb value:
[tex]8.50* 10^{-7} = \frac{(2.3988* 10^{-4})^{2} }{x-2.3988*10^{-4} } [/tex]

Solve the equation algebraically to get x.
X = 6.8*10^-2 

Hope that helps. 
If you need more clarification just let me know

ardni313

Concentration of  weak base : 0.0679

Further explanation

pH is the degree of acidity of a solution that depends on the concentration of H ions. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.

A solution whose value is different by n has a difference in the concentration of H ions of 10ⁿ.

According to Arrhenius, Base is a compound which in water releases OH-ions

Electrolytes (acids or bases) with a small degree of ionization are classified as weak electrolytes

OH concentration and molarity can be related to the formula

[tex]\rm [OH^-] = \sqrt{Kb.M}[/tex]

Kb = base ionization constant

M = molarity

A certain weak base has a Kb of 8.50 × 10⁻⁷

pH = 10.38

pOH = 14 - 10.38

pOH = 3.62

[tex]\rm [OH^-]=10^{-3.62}[/tex]

Then :

[tex]\rm [10^{-3.62}^2] = \sqrt{8.5.10^{-7}.M}[/tex]

[tex]\rm M=\dfrac{10^{-7.24}}{8.5.10^{-7}}\\\\M=1.18.10^{-1.24}\\\\M=\boxed{\bold{0.0679}}[/tex]

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