What is the final concentration (M) of a solution prepared by diluting 50.0 mL of a solution to a volume of

(3) A 10.00-mL sample of 0.1000 M KH2PO4 was titrated with 0.1000 M HCl Ka for phosphoric acid (H3PO4): Ka1= 7.50x10-3; Ka2=6.20

shtirl [24]

Answer:

The pH of this solution is 1,350

Explanation:

The phosphoric acid (H₃PO₄) has three acid dissociation constants:

HPO₄²⁻ ⇄ PO4³⁻ + H⁺ Kₐ₃ = 4,20x10⁻¹⁰ (1)

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸ (2)

H₃PO₄ ⇄ H₂PO4⁻ + H⁺ Kₐ₁ = 7,50x10⁻³ (3)

The problem says that you have 10,00 mL of KH₂PO₄ (It means H₂PO₄⁻) 0,1000 M and you add 10,00 mL of HCl (Source of H⁺) 0,1000 M. So you can see that we have the reactives of the equation (3).

We need to know what is the concentration of H⁺ for calculate the pH.

The moles of H₂PO₄⁻ are:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

The moles of H⁺ are, in the same way:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

So:

H₃PO₄ ⇄ H₂PO4⁻ + H⁺ Kₐ₁ = 7,50x10⁻³ (3)

X mol ⇄ (1x10⁻³-X) mol + (1x10⁻³-X) mol (4)

The chemical equilibrium equation is:

Kₐ₁ = ([H₂PO4⁻] × [H⁺] / [H₃PO₄]

So:

7,50x10⁻³ = (1x10⁻³-X)² / X

Solving the equation you will obtain:

X² - 9,5x10⁻³ X + 1x10⁻⁶ = 0

Solving the quadratic formula you obtain two roots:

X = 9,393x10⁻³ ⇒ This one has no chemical logic because solving (4) you will obtain negative H₂PO4⁻ and H⁺ moles

X = 1,065x10⁻⁴

So the moles of H⁺ are : 1x10⁻³- 1,065x10⁻⁴ : 8,935x10⁻⁴ mol

The reaction volume are 20,00 mL (10,00 from both KH₂PO₄ and HCL)

Thus, the molarity of H⁺ ([H⁺]) is: 8,935x10⁻⁴ mol / 0,02000 L = 4,468x10⁻² M

pH is -log [H⁺]. So the obtained pH is 1,350

I hope it helps!

5 0

3 years ago

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Agata [3.3K]

<u>Answer:</u> The molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

$PbI_2\rightleftharpoons Pb^{2+}+2I^-$

s 2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Pb^{2+}][I^-]^2$

We are given:

$K_{sp}=7.9\times 10^{-9}$

Putting values in above equation, we get:

$7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L$

Hence, the molar solubility of $PbI_2$ is $1.25\times 10^{-3}mol/L$

3 0

3 years ago

Which compound is an exception to the octet rule? H20 HCI CCI4 CIF3

Reika [66]

Answer:

clf3

Explanation:

it occupied more than 8 valence electrons

8 0

3 years ago

in this cut-away image of an atom, where is the red arrow most likely pointed at the greatest probability of locating an electro

emmasim [6.3K]

Answer:

c

Explanation:

3 0

3 years ago

Read 2 more answers

A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25

mylen [45]

<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

Concentration of stock solution as 1.00 M
Volume of the stock solution as 50 mL
Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

= (1.00 M × 50 mL) ÷ 0.25 M

= 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

5 0

3 years ago