using NaH2PO4 and Na2HPO4 both are stock solutions of .2 M 3) Calculate volumes of both...

using NaH2PO4 and Na2HPO4 both are stock solutions of .2 M

3) Calculate volumes of both solutions required to prepare 150 mL 0.1 M phosphate buffer

given pKa 6.8
and we want the pH to be 7.0

im not sure if you need the gram values from the preceding problems for 50 ml solutions.. we went over this today the formula was something like
(x/.1-x)=10^7-6.8
then somehow the value of mLs was calculated and a part of the solution was with water.. Thanks

Solutions

Expert Solution

Answer – We are given pKa = 6.8 and pH = 7.0

Total volume = 150 mL, phosphate buffer = 0.1 M

We know Henderson–Hasselbalch equation

pH = pKa + log [Conjugate base] / [Acid]

7.0 = 6.8 + log [HPO₄^2-] / [H₂PO₄^-]

7.0 -6.8 = log [HPO₄^2-] / [H₂PO₄^-]

0.2 = log [HPO₄^2-] / [H₂PO₄^-]

Taking antilog from both side

[HPO₄^2-] / [H₂PO₄^-] = 1.58

[HPO₄^2-] = 1.58 * [H₂PO₄^-]

We know total volume is 150 , so volume of Na₂HPO₄ = 150 – x

Moles of Na₂HPO₄ = volume (L) * molarity

= 1.58 * x/1000 * 0.1 M

150-x/1000 *0.1 = 1.58 * x/1000 * 0.1 M

150-x /1000 = 1.58x / 1000

150-x = 1.58x

150 = 1.58x +x

150 = 2.58x

So, x = 150 /2.58

= 58.14 mL

Volume of NaH₂PO₄ = 58.14 mL

Volume of Na₂HPO₄ = 150 – x

= 150 -58.14

= 91.86 mL

queen_honey_blossom answered 1 year ago

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