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geniusboy [140]
3 years ago
13

For Na2HPO4:(( (Note that for H3PO4, ka1= 6.9x10-3, ka2 = 6.4x10-8, ka3 = 4.8x10-13 ) a) The active anion is H2PO4- b) The activ

e anion is HPO4 3- c) is basic d) Is acidic
Chemistry
1 answer:
Komok [63]3 years ago
6 0

Answer:

Check the explanation

Explanation:

Answer – Given, H_3PO_4 acid and there are three Ka values

K_{a1}=6.9x10^8, K_{a2} = 6.2X10^8, and K_{a3}=4.8X10^{13}

The transformation of H_2PO_4- (aq) to HPO_4^2-(aq)is the second dissociation, so we need to use the Ka2 = 6.2x10-8 in the Henderson-Hasselbalch equation.

Mass of KH2PO4 = 22.0 g , mass of Na2HPO4 = 32.0 g , volume = 1.00 L

First we need to calculate moles of each

Moles of KH2PO4 = 22.0 g / 136.08 g.mol-1

                             = 0.162 moles

Moles of Na2HPO4 = 32.0 g /141.96 g.mol-1

                             = 0.225 moles

[H2PO4-] = 0.162 moles / 1.00 L = 0.162 M

[HPO42-] = 0.225 moles / 1.00 L = 0.225 M

Now we need to calculate the pKa2

pKa2 = -log Ka

       = -log 6.2x10-8

       = 7.21

We know Henderson-Hasselbalch equation

pH = pKa + log [conjugate base] / [acid]

pH = 7.21 + log 0.225 / 0.162

     = 7.35

The pH of a buffer solution obtained by dissolving 22.0 g of KH2PO4 and 32.0 g of Na2HPO4 in water and then diluting to 1.00 L is 7.35

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<h3>F=4k.gm/s^2</h3>

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3 years ago
Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water, determine the const
shtirl [24]

Answer:

  • m = 1,000/58.5
  • b = - 1,000 / 58.5

1) Variables

  • molarity: M
  • density of the solution: d
  • moles of NaCl: n₁
  • mass of NaCl: m₁
  • molar mass of NaCl: MM₁
  • total volume in liters: Vt
  • Volume of water in mililiters: V₂
  • mass of water: m₂

2) Density of the solution: mass in grams / volume in mililiters

  • d = [m₁ + m₂] / (1000Vt)

3) Mass of NaCl: m₁

    Number of moles = mass in grams / molar mass

    ⇒ mass in grams = number of moles × molar mass

        m₁ = n₁ × MM₁


4) Number of moles of NaCl: n₁

   Molarity = number of moles / Volume of solution in liters

   M = n₁ / Vt

   ⇒ n₁ = M × Vt


5) Substitue in the equation of m₁:

   m₁ = M × Vt × MM₁


6) Substitute in the equation of density:

    d = [M × Vt × MM₁ + m₂] / (1000Vt)


7) Simplify and solve for M

  • d = M × Vt × MM₁ / (1000Vt) + m₂/ (1000Vt)
  • d = M × MM₁ / (1000) + m₂/ (1000Vt)

Making the simplistic assumption that the dissolved NaCl(s) does not affect the volume of the solvent water means 1000Vt = V₂  

  • d = M × MM₁ / (1000) + m₂/ V₂

        m₂/ V₂ is the density of water: 1.00 g/mL

  • d = M × MM₁ / (1000) + 1.00 g/mL
  • M × MM₁ / (1000) = d - 1.00 g/mL
  • M = [1,000/MM₁] d - 1,000/ MM₁

8) Substituting MM₁ = 58.5 g/mol

  • M = [1,000/58.5] d - [1,000/ 58.5]

Comparing with the equation Molarity = m×density + b, you obtain:

  • m = 1,000/58.5
  • b = - 1,000/58.5
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sasho [114]
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