Respuesta :
Answer :
The correct answer for Mass of Na₂HPO₄ = 4.457 g and mass of NaH₂PO₄ = 8.23 g
Given : pH = 6.86
Total concentration of Phosphate buffer = 0.1 M
Asked : Mass of Sodium phosphate monobasic (NaH₂PO₄) = ?
Mass of Sodium phosphate dibasic(Na₂HPO₄)= ?
Following steps can be done to find the masses of NaH₂PO₄ and Na₂HPO₄ :
(In phosphate buffer , Na+ ion from NaH₂PO₄ and Na₂HPO₄ acts as spectator ion , so only H₂PO₄⁻ and HPO₄²⁻ will be considered )
Step 1 : To find pka
H₂PO₄⁻ <=> HPO₄²⁻
The above reaction has pka = 7.2 ( from image shown )
Step 2 : Plug values in Hasselbalch- Henderson equation .
Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :
[tex]pH = pka + log\frac{[A^-]}{[HA]}[/tex]
pH = 6.86 pKa = 7.2
[tex]6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}[/tex]
Subtracting both side by 7.2
[tex]6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}[/tex]
[tex]-0.34 = log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}[/tex]
Removing log
[tex]10^-^0^.^3^4 = \frac{[HPO_4^2^-]}{[H_2PO_4^-]}[/tex]
[tex]\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457[/tex] ---------------- equation (1)
Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻
Total concentration of buffer = [H₂PO₄⁻] + [HPO₄²⁻] = 0.1 M
Hence, [H₂PO₄⁻ ] + [ HPO₄²⁻ ] = 0.1 M
Assume [H₂PO₄⁻ ] = x
So , [x ] + [ HPO₄²⁻ ] = 0.1 M
[ HPO₄²⁻ ] = 0.1 - x
Step 4 : Plugging value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ]
[H₂PO₄⁻ ] = x
[ HPO₄²⁻ ] = 0.1 - x
Equation (1) = >[tex]\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457[/tex]
Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :
[tex]\frac{[0.1 - x ]}{[ x]} = 0.457[/tex]
Cross multiplying
[tex]0.1 - x = 0.457 x[/tex]
Adding x on both side
[tex]0.1 -x + x = 0.457 x + x[/tex]
[tex]0.1 = 1.457 x[/tex]
Dividing both side by 1.457
[tex]\frac{0.1}{1.457} = \frac{1.457 x }{1.457}[/tex]
x = 0.0686 M
Hence , [H₂PO₄⁻ ] = x = 0.0686 M
[ HPO₄²⁻ ] = 0.1 - x
[ HPO₄²⁻ ] = 0.1 - 0.0686
[ HPO₄²⁻ ] = 0.0314 M
Step 5 : To find moles of H₂PO₄⁻ ( NaH₂PO₄) and HPO₄²⁻ (Na₂HPO₄ ) .
Molarity is defined as mole of solute per 1 L volume of solution .
Molarity of NaH₂PO₄ = 0.0686 M or 0.0686 mole per 1 L
Molarity of Na₂HPO₄ = 0.0314 M or 0.0314 mole per 1 L
Since that volume of buffer solution is 1 L , so Molarity = mole
Hence Mole of NaH₂PO₄ = 0.0686 mol
Mole of Na₂HPO₄ = 0.0314 mol
Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄
Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :
Molar mass of Na₂HPO₄ = [tex]141.96 \frac{g}{mol}[/tex]
Molar mass of NaH₂PO₄ = [tex]119.98 \frac{g}{mol}[/tex]
[tex]Mass (g) = mole (mol)* molar mass(\frac{g}{mol})[/tex]
[tex]Mass of Na_2HPO_4 = 0.0314 mol * 141.96 \frac{g}{mol}[/tex]
Mass of Na₂HPO₄ = 4.457 g
[tex]Mass of NaH_2PO_4 = 0.0686 mol * 119.98 \frac{g}{mol}[/tex]
Mass of NaH₂PO₄ = 8.23 g
