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 )
<u>Step 1 : To find pka </u>
H₂PO₄⁻ <=> HPO₄²⁻
The above reaction has pka = 7.2 ( from image shown )
<u>Step 2 : Plug values in Hasselbalch- Henderson equation </u>.
Hasselbalch -Henderson equation is to find pH for buffer solution which is as follows :
![pH = pka + log\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=pH%20%3D%20pka%20%2B%20log%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
pH = 6.86 pKa = 7.2
![6.86 = 7.2 + log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}](https://tex.z-dn.net/?f=6.86%20%3D%207.2%20%2B%20log%20%5Cfrac%7B%5BHPO_4%5E2%5E-%5D%7D%7B%5BH_2PO_4%5E-%5D%7D)
Subtracting both side by 7.2
![6.86-7.2 = 7.2 -7.2+ log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}](https://tex.z-dn.net/?f=6.86-7.2%20%3D%207.2%20-7.2%2B%20log%20%5Cfrac%7B%5BHPO_4%5E2%5E-%5D%7D%7B%5BH_2PO_4%5E-%5D%7D)
![-0.34 = log \frac{[HPO_4^2^-]}{[H_2PO_4^-]}](https://tex.z-dn.net/?f=-0.34%20%3D%20%20log%20%5Cfrac%7B%5BHPO_4%5E2%5E-%5D%7D%7B%5BH_2PO_4%5E-%5D%7D)
Removing log
![10^-^0^.^3^4 = \frac{[HPO_4^2^-]}{[H_2PO_4^-]}](https://tex.z-dn.net/?f=10%5E-%5E0%5E.%5E3%5E4%20%3D%20%20%20%5Cfrac%7B%5BHPO_4%5E2%5E-%5D%7D%7B%5BH_2PO_4%5E-%5D%7D)
---------------- equation (1)
<u>Step 3 : To find molarity of H₂PO₄⁻ and HPO₄²⁻</u>
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) = >![\frac{[HPO_4^2^-]}{[ H_2PO_4^-]} = 0.457](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO_4%5E2%5E-%5D%7D%7B%5B%20H_2PO_4%5E-%5D%7D%20%3D%200.457)
Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :
![\frac{[0.1 - x ]}{[ x]} = 0.457](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0.1%20-%20x%20%5D%7D%7B%5B%20x%5D%7D%20%3D%200.457)
Cross multiplying

Adding x on both side


Dividing both side by 1.457

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
<u>Step 6 : To find mass of Na₂HPO₄ and NaH₂PO₄ </u>
Moles of Na₂HPO₄ and NaH₂PO₄ can be converted to their masses using molar mass as follows :
Molar mass of Na₂HPO₄ = 
Molar mass of NaH₂PO₄ = 


Mass of Na₂HPO₄ = 4.457 g

Mass of NaH₂PO₄ = 8.23 g