Science question down below

Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81

Gnom [1K]

Answer:

Explanation:

Given parameters;

pH = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

pH = -log[H₃O⁺]

pOH = -log[OH⁻]

pH + pOH = 14

Now let us solve;

pH = 8.74

since pH = -log[H₃O⁺]

8.74 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{8.74}$

[H₃O⁺] = 1.82 x 10⁻⁹mol dm³

pH + pOH = 14

pOH = 14 - 8.74

pOH = 5.26

pOH = -log[OH⁻]

5.26 = -log[OH⁻]

[OH⁻] = 10 $^{-5.26}$

[OH⁻] = 5.5 x 10⁻⁶mol dm³

2. pH = 11.38

since pH = -log[H₃O⁺]

11.38 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{11.38}$

[H₃O⁺] = 4.17 x 10⁻¹² mol dm³

pH + pOH = 14

pOH = 14 - 11.38

pOH = 2.62

pOH = -log[OH⁻]

2.62 = -log[OH⁻]

[OH⁻] = 10 $^{-2.62}$

[OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

since pH = -log[H₃O⁺]

2.81 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{2.81}$

[H₃O⁺] = 1.55 x 10⁻³ mol dm³

pH + pOH = 14

pOH = 14 - 2.81

pOH = 11.19

pOH = -log[OH⁻]

11.19 = -log[OH⁻]

[OH⁻] = 10 $^{-11.19}$

[OH⁻] =6.46 x 10⁻¹²mol dm³

5 0

3 years ago

Which direction does the sun to move across the sky

Ierofanga [76]

Answer:

EAST is the answer

Explanation:

For the moment, let us just think about one motion - - Earth's spin (or rotation) on its axis. Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

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2 years ago

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if a scientist creates a new rock like substance in a laboratory, why wouldnt this type of material be classified as a true mine

ad-work [718]

The reason why it is not considered this is because the material was made in a lab, not through nature, which is what is required to be considered as a true mineral.

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2 years ago

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A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

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At what temperature <br> does water boil

Ipatiy [6.2K]

The temperature is a 100

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