Units used to measure the rate of electron flow

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

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

(c) a solution with a pOH of 12.

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)

3 0

3 years ago

When an electron of a hydrogen atom travels from n = 4 to n = 2, a photon of energy 4.165 × 10-19 joules is released. Calculate

gulaghasi [49]

Answer:

Explanation:

please delete this it was a mistake

4 0

3 years ago

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Which of the following claims may be made based on your observations? (You will recieve negative points for wrong answers.)

Ostrovityanka [42]

Answer:

D, E and F

Explanation:

About tetrachloro cobalt complexes, the following facts have been observed

Color of the tetrachloro cobalt complexes is blue.
They do not decompose on heating that means synthesis of tetra chloro is endothermic.

About hexa aqua cobalt complexes, the following facts have been observed

Color of the hexa aqua cobalt complexes is pink color.
They decompose on heating and remain stable on cooling that means process of synthesis of hexa aqua cobalt complexes is exothermic.

Based on above, the correct statements are:

The correct is chloro cobalt complex is blue and aqua cobalt

complex is pink.

The chloro complex is favored by heating.

If the chloro complex is a product, then the reaction must be endothermic.

The correct options are D, E and F.

8 0

3 years ago

ΔH for the formation of CuCl2 from its elements is -220.1 kJ/mol. How many kJ are associated with the formation of 0.30 mole of

otez555 [7]

0.3 * - 220.1 = - 66.03kJ

3 0

3 years ago

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What is the color of benzene and bromine

Harman [31]

Bromine is a reddish-brown color.
Benzene is clear or colorless

6 0

4 years ago

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