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Calculate the ph of a buffer solution obtained by dissolving 13.0 g of kh2po4(s) and 26.0 g of na2hpo4(s) in water and then diluting to 1.00 l.

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superman1987
To get the PH we are going to use Henderson - Hasselblach equation:

PH = Pka + ㏒ [A/AH]

when the molar mass of Na2HPO4 = 142 g/mol 

and A is the conjugate base HPO4-- ions so,

∴[A] = 26g / 142 g/mol 
        = 0.183 M
and when the molar mass of KH2PO4 = 136 g/mol

and AH is the weak acid H2PO4- ions so,

∴[AH] = 13 g / 136 g/mol 
          = 0.096 M

and when we have the Pka value of H3PO4 = 7.21 

so, by substitution:

∴ PH = 7.21 + ㏒ (0.183 / 0.096)
         = 7.49
BladeRunner212

7.49

Further explanation

Given:

  • A buffer solution obtained by dissolving 13.0 g of KH₂PO₄ and 26.0 g of Na₂HPO₄ in water and then diluting to 1.00 L.
  • pKa = 7.21.

Question:

Calculate the pH of this buffer.

The Process:

Step-1

Let us first observe the ionization reaction of the Na₂HPO₄ salt below.

 [tex]\boxed{ \ Na_2HPO_4 \rightleftharpoons 2Na^+ + HPO_4^{2-} \ }[/tex]

  • The Na₂HPO₄ salt has valence = 1 according to the number of HPO₄²⁻ ions as a weak part.
  • H₂PO₄⁻ and HPO₄²⁻ are conjugate acid-base pairs .
  • KH₂PO₄ and Na₂HPO₄ form an acidic buffer system.

Step-2

Let us prepare the mole number.

  • The molar mass of KH₂PO₄ is Mr = 136 g/mol
  • The molar mass of Na₂HPO₄ is Mr = 142 g/mol

[tex]\boxed{ \ n = \frac{mass}{Mr} \ }[/tex]

  • KH₂PO₄ ⇒ [tex]\boxed{ \ \frac{13.0 \ g}{136 \ g/mol} = 0.096 \ moles \ }[/tex]
  • Na₂HPO₄ ⇒ [tex]\boxed{ \ \frac{26.0 \ g}{142 \ g/mol} = 0.183 \ moles \ }[/tex]

Step-3

To calculate the specific pH of a given buffer, we need using The Henderson-Hasselbalch equation for acidic buffers:  

 [tex]\boxed{ \ pH = pK_a + log\frac{[A^-]}{[HA]} \ }[/tex]

where,  

  • Ka represents the dissociation constant for the weak acid;  
  • [A-] represent the concentration of the conjugate base (i.e. salt);  
  • [HA] is the concentration of the weak acid.

Here we don't need to calculate both molarities because they dissolve in the same amount of volume. Or, if you want to keep finding out, then:

[tex]\boxed{ \ [HPO_4^{2-}] = \frac{0.183 \ moles}{1.00 \ L} = 0.183 \ M \ }[/tex]

[tex]\boxed{ \ [H2PO_4^{-}] = \frac{0.096 \ moles}{1.00 \ L} = 0.096 \ M \ }[/tex]

So, let us calculate the pH now.

[tex]\boxed{ \ pH = pK_a + log \frac{[HPO_4^{2-}]}{[H_2PO_4^-]} \ }[/tex]

 [tex]\boxed{ \ pH = 7.21 + log \frac{[0.183]}{[0.096]} \ }[/tex]

[tex]\boxed{ \ pH = 7.21 + 0.28 \ }[/tex]

Thus, the pH of a buffer solution equal to [tex]\boxed{\boxed{ \ pH = 7.49 \ }}[/tex]

Learn more

  1. Calculate the pH of the buffer HC₂H₃O₂ and KC₂H₃O₂. https://brainly.com/question/9081563
  2. What would be the final ph if 0.0100 moles of solid NaOH were added to this buffer? https://brainly.com/question/9082442  
  3. Calculate the change in pH when 2.0 x 10⁻² mol of NaOH is added to this buffer? https://brainly.com/question/3446303  

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