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2 years ago

Suggest the pH of the solution made when potassium oxide dissolves in water.

Chemistry

2 answers:

Naddika [18.5K]2 years ago

8 0

Potassium oxide (K₂O): ionic compound.

Reaction:

K₂O + H₂O →2KOH

KOH = strong base, pH>7

Varvara68 [4.7K]2 years ago

4 0

Answer:

Since KOH is a strong base, the solution completely ionizes into K+ and OH- when in water. The reaction KOH --> K+ + OH- takes place. The concentration of [ OH- ] can then be used to calculate the pOH of the solution. pH = 14 - pOH = 14 - 1.48 = 12.52

Explanation:

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You decided to prepare a phosphate buffer from solid sodium dihydrogen phosphate (NaH2PO4) and disodium hydrogen phosphate (Na2H

KIM [24]

Answer:

For disodium hydrogen phosphate:

5.32g Na2HPO4

For sodium dihydrogen phosphate:

7.65g Na2HPO4

Explanation:

First, you have to put all the data from the problem that you going to use:

-NaH2PO4 (weak acid)

-Na2HPO4 (a weak base)

-Volume = 1L

-Buffer pH = 7.00

-Concentration of [NaH2PO4 + Na2HPO4] = 0.100 M

What we need to find the pKa of the weak acid, in this case NaH2PO4, for that you need to find the Ka (acid constant) of NaH2PO4, and for this we use the pKa of the phosphoric acid as follow:

H3PO4 = H2PO4 + H+ pKa1 = 2.14

H2PO4 = HPO4 + H+ pKa2 = 6.86

HPO4 = PO4 + H+ pKa3 = 12.4

So, for the preparation of buffer, you need to use the pKa that is near to the value of the pH that you want, so the choice will be:

pKa2= 6.86

Now we going to use the Henderson Hasselbalch equation for the pH of a buffer solution:

pH = pKa2 + log [(NaH2PO4)/(Na2HPO4)]

The solution of the problem is attached to this answer.

Download odt

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3 years ago

A chef plans to mix 100% vinegar with italian dressing. the italian dressing contains 13% vinegar. the chef wants to make 150 mi

charle [14.2K]

To determine the volume of both concentration of vinegar, we need to set up two equations since we have two unknowns.

For the first equation, we do a mass balance:

mass of 100% vinegar + mass of 13% vinegar = mass of 42% vinegar

Assuming they have the same densities, then we can write this equation in terms of volume.

V(100%) + V(13%) = V(42%)
we let x = V(100%)
y = V(13%)

x + y = 150

For the second equation, we do a component balance:

1.00x + .13y = 150(.42)
x + .13y = 63

The two equations are
x + y = 150
x + .13y = 63

Solving for x and y,
x = 50
y = 100

Therefore, you need to mix 50 mL of the 100% vinegar and 100 mL of the 13% vinegar.

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Answer:

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Explanation:

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Answer:

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Explanation:

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