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IgorLugansk [536]
3 years ago
15

What is the pH of this solution?

Chemistry
1 answer:
Vesnalui [34]3 years ago
5 0

Answer:

pH = 11.216.

Explanation:

Hello there!

In this case, according to the ionization of ammonia in aqueous solution:

NH_3+H_2O\rightleftharpoons NH_4^++OH^-

We can set up its equilibrium expression in terms of x as the reaction extent equal to the concentration of each product at equilibrium:

Kb=\frac{[NH_4^+][OH^-]}{[NH_3]} \\\\1.80x10^{-5}=\frac{x*x}{0.150-x}

However, since Kb<<<1 we can neglect the x on bottom and easily compute it via:

1.80x10^{-5}=\frac{x*x}{0.150}\\\\x=\sqrt{1.80x10^{-5}*0.150}=1.643x10^{-3}M

Which is also:

[OH^-]=1.643x10^{-3}M

Thereafter we can compute the pOH first:

pOH=-log(1.643x10^{-3}M)\\\\pOH=2.784

Finally, the pH turns out:

pH=14-2.784\\\\pH=11.216

Regards!

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Hey buddy I am here to help!

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