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Shalnov [3]
3 years ago
15

Calculate the molarity of a solution of sodium hydroxide, naoh, if 23.64 ml of this solution is needed to neutralize 0.5632 g of

potassium hydrogen phthalate, khc8h4o4.
Chemistry
1 answer:
GuDViN [60]3 years ago
8 0
The  molarity  of  NaOH  needed  is  calculated   as  follows
calculate  the  moles  of  KhC8h4O4

that  is  moles  =  mass/molar  mass  of  KhC8h4O4(204.22 g/mol)

=0.5632g /204.22g/mol=  2.76  x10^-3  moles

write the  equation  for  reaction

khc8h4O4  +  NaOH  ---> KNaC8h4O4  +  H2O

from  the  equation  above  the   reacting  ratio   of   KhC8h4O4  to  NaOh  is  1:1  therefore  the  moles  of  Naoh  is  also  2.76  x10^-3  moles

molarity  of NaOh  =  (moles  of  NaOh /  volume ) x  1000

that  is { (2.76  x10^-3) / 23.64}  x100  =0.117 M
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Two children on roller skates stand facing each other. The child on the right puts her arms out and pushes away from her partner
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They both go backward because of force.

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4. Energy can be conserved by -
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Energy can be conserved by efficient energy use.

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4 0
3 years ago
(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
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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!

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