using the Henderson hasselbalch equation , calculate the g of NaH2PO4*H2O and g

Question

using the Henderson hasselbalch equation , calculate the g of NaH2PO4*H2O and g of Na2HPO4 needed to prepare 100 ml of 100 mM sodium phosphate buffer pH 7.0.
The pka value should be 6.82 but I could be wrong. Please show all steps!! using the Henderson hasselbalch equation , calculate the g of NaH2PO4*H2O and g of Na2HPO4 needed to prepare 100 ml of 100 mM sodium phosphate buffer pH 7.0.
The pka value should be 6.82 but I could be wrong. Please show all steps!! using the Henderson hasselbalch equation , calculate the g of NaH2PO4*H2O and g of Na2HPO4 needed to prepare 100 ml of 100 mM sodium phosphate buffer pH 7.0.
The pka value should be 6.82 but I could be wrong. Please show all steps!!

Explanation / Answer

We know the Henderson hasselbalch equation is as below

pH = pKa + log (base/acid)

The pKa for the dissociation of ionization hydrogen of dihydrogen phosphate is 6.86.

H2PO4- <-----> HPO42- + H+

Now pH = pKa + log (base/acid)

        7.0 = 6.86 + log (base/acid)

        log (base/acid) = 7.0 - 6.86

         log (base/acid) = 0.14

        (base/acid) = 100.14   = 1.38

This ratio is approx 7/5; that is 7 parts Na2HPO4 to 5 parts NaH2PO4*H2O

Beacase [ HPO42-] + [H2PO4- ] = 0.100 M , [H2PO4- ] = 0.100M - [ HPO42-] { here 100 mM = 0.100 M }

[ HPO42-]   / {0.100M - [ HPO42-]} =1.38

[ HPO42-] = 0.138 / 2.38 M = 0.058 M = 0.058 mol / L

[H2PO4- ] = 0.100M - [ HPO42-]   = 0.100M - 0.058 M = 0.042 mol /L

The amound needed for 100 ml of solution = M.W x (mol/L)

For NaHPO4*H2O = (138.01 g/mol) x (0.042 mol /L ) x 0.1 = 0.579 gram

For Na2HPO4 = (141.98 g/mol) x (0.058 mol /L ) x 0.1 = 0.823 gram

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