Chemistry, asked by sixsi, 5 months ago

What volume of 0.841 M KOH solution is required to make 3.53 L of a solution with pH of 12.80?

Answers

Answered by AkashMathematics
0

Answer:

Explanation:

Start by using the

pOH

of the solution to figure out the concentration of hydroxide anions,

OH

. You should know that

pOH

=

log

(

[

OH

]

)

−−−−−−−−−−−−−−−−−−−−

To solve for the concentration of hydroxide anions, rearrange the equation as

log

(

[

OH

]

)

=

pOH

This can be written as

10

log

(

[

OH

]

)

=

10

pOH

which is equivalent to

[

OH

]

=

10

pOH

Now, an aqueous solution at room temperature has

pH

+

pOH

=

14

−−−−−−−−−−−−−−−−

This means that the

pOH

of the target solution is equal to

pOH

=

14

12.4

=

1.6

Consequently, the target solution must have

[

OH

]

=

10

1.6

=

2.51

10

2

M

As you know, molarity is defined as the number of moles of solute present in

1 L

of solution. This means that the number of moles of potassium hydroxide needed to ensure that concentration of hydroxide anions is equal to

3.55

L solution

2.51

10

2

moles OH

1

L solution

=

8.91

10

2

moles OH

All you have to do now is figure out what volume of the stock solution contains the needed number of moles of hydroxide anions

8.91

10

2

moles OH

1 L solution

0.855

moles OH

=

0.104 L

Expressed in milliliters, the answer will be

V

stock KOH

=

104 mL

−−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three sig figs.

So, in order to prepare this solution, take

104 mL

of

0.855 M

stock potassium hydroxide solution and add enough water until the final volume of the solution is equal to

3.55 L

.

Similar questions