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frutty [35]
3 years ago
8

Calculate the pH of a 0.187 M solution of a base that has a Kb = 5.86x10^-6

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
5 0

Answer:

pH=11

Explanation:

Hello there!

In this case, since the ionization of the described base is:

base+H_2O\rightarrow OH^-+baseH^+

Thus, the equilibrium expression in terms of the reaction extent is:

Kb=5.86x10^{-6}=\frac{x^2}{0.187-x}

Thus, by solving for x we obtain:

5.86x10^{-6}({0.187-x})=x^2\\\\1.10x10^{-6}-5.86x10^{-6}x-x^2=0

So the values of x are:

x_1=0.0010M\\x_2=-0.0010M

So the feasible answer is 0.0010 M, thus we compute the pOH:

pOH=-log(0.0010M)=3

And therefore the pH:

pH=14-pOH=14-3\\\\pH=11

Best regards!

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