tankhill5534

09.04.2021 •

Chemistry

If I have a sealed tank of oxygen at 85 degrees celsius and I cool it down to 50 degrees celsius. What happens to the pressure? *

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Ответ:

mya1318

23.01.2023

Chemistry

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

Now we have to calculate the OH^- concentration.

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

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the OH^- concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M NH_3

As we know that 1 mole of NH_3 is a weak base. So, in a solution it will not dissociates completely.

So, the OH^- concentration will be less than 0.10 M

We have to calculate the OH^- concentration.

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

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the OH^- concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that NaOH is a strong base. So, it dissociates to give Na^+ ion and OH^- ion.

So, 0.10 M of NaOH in a solution dissociates to give 0.10 M of Na^+ ion and 0.10 M of OH^- ion.

Thus, the OH^- concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of HNO_2

As we know that 1 mole of HNO_2 in a solution dissociates to give 1 mole of H^+ ion and 1 mole of NO_2^- ion.

So, $1\times 10^{-4}M$ of HNO_2 in a solution dissociates to give $1\times 10^{-4}M$ of H^+ ion and $1\times 10^{-4}M$ of NO_2^- ion.

The concentration of H^+ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

pH=4

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

Now we have to calculate the OH^- concentration.

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

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the OH^- concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

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