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Contact [7]
2 years ago
7

What is the pH of a solution of RbOH with a concentration of 0.86 M? Answer to 2 decimal places

Chemistry
1 answer:
lubasha [3.4K]2 years ago
3 0

Answer:The pH of the solution is given by pH=−log([H3O+])

Explanation:so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−

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Explanation:

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The quantum number of an electron consists of four number:

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  • <em>l</em> (subshell number or  orbital number, 0 - orbital <em>s</em>, 1 - orbital <em>p</em>, 2 - orbital <em>d...</em>)
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  • <em>n </em>= 2, shell number 2,
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  • m_{l} = 0, 2nd "box" in the range -1, 0, 1
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