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4 years ago

Who wants to be my friend, and I just began

Chemistry

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shepuryov [24]4 years ago

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Answer:

i want to be your friend

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3 years ago

Calculate the pH of a 0.2 M * 4 solution for which Kb = 1.8*10^-5 at 26 c . The equation for the reaction

lara31 [8.8K]

<u>Answer:</u> The pH of the solution is 11.24

<u>Explanation:</u>

We are given:

Molarity of ammonia = 0.2 M

$K_b=1.8\times 10^{-5}$

The given chemical equation follows:

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

I: 0.2

C: -x +x +x

E: 0.2-x x x

The expression for equilibrium constant follows:

$K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}$

Putting values in above expression, we get:

$1.8\times 10^{-5}=\frac{x^2}{0.2-x}\\\\1.8\times 10^{-5}(0.2-x)=x^2\\\\x^2+(1.8\times 10^{-5}x)-(0.36\times 10^{-5})=0\\\\x=1.88\times 10^{-3}, 1.9\times 10^{-3}$

Neglecting the negative value of x as concentration cannot be negative.

So, $[OH^-]=x=1.88\times 10^{-3}M$

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.

$pOH=-\log [OH^-]$

Putting values in above equation, we get:

$pOH=-\log (1.88\times 10^{-3})\\\\pOH=2.76$

We know:

$pH+pOH=14\\\\pH=14-2.76\\\\pH=11.24$

Hence, the pH of the solution is 11.24

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C3H8+5O2 --> 3CO2 + 4H2O

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Explanation:

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