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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 = 6.86 pKa = 7.2
Subtracting both side by 7.2
Removing log
---------------- 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) = >
Plug value of [H₂PO₄⁻ ] and [ HPO₄²⁻ ] ( from step 3 ) into equation (1) as :
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
