I need help with the amount of grams of each I would need. If you can walk me th

Question

I need help with the amount of grams of each I would need. If you can walk me through this step by step it would be appreciated. Thanks!

Phosphoric acid is a triprotic acid with the following pKa values: pK -2.148 pK 7.198 pK-12375 You wish to prepare 1.000 L of a 0.0200 M phosphate buffer at pH 7.420. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. A4 of k. (click to edit 1 of each salt will you add to the mixture? Number Number mass NaH,PO,963g 0,-0.963 mass Na,HPO1.7 What other combination of phosphoric acid and/or its salts could be mixed to prepare this buffer? (Check all that apply). HPO4 and NaH2PO4 H3PO4 and Na2HPOs H3PO4 and NaPO4 NaH2PO4 and NasPO Na2HPO4 and Napo

Explanation / Answer

the pKa2 is for : (H2PO4) <=> H+ & (HPO4)

The desired pH is essentially that fo Ka2's pKa: 7.420 vs. pKa2'a : 7.198

pH = pKa + log [(HPO4-)] / [(H2PO4)]

7.42= 7.198 + log [(HPO4-)] / [(H2PO4)]

7.42 - 7.198 = log [(HPO4-)] / [(H2PO4)]

0.222 = log [(HPO4-)] / [(H2PO4)]
taking antilog
1.667 = [(HPO4-)] / [(H2PO4)]
so
[(HPO4-)] = 1.667 of the strength of [(H2PO4)]

Let the strenght of HPO4- = y

we find that a 0.0200 M phosphate buffer = y+ [1.667 y]

so
2.667 y = 0.020 Molar

on solving for y
y, the strength of [(H2PO4)], = 0.00749 Molar NaH2PO4

& solving for [1.667 y], the strength of [(HPO4-)], = 0.0124 Molar Na2HPO4

solving for moles in a 1.000 Litre buffer:
0.00749 moles NaH2PO4

0.00124 moles Na2HPO4
as we have moles we can solve for mass by using molar mass

(0.00749 moles NaH2PO4) (119.98 grams per mole) = 0.898 grams of NaH2PO4

(0.00124 moles Na2HPO4) ( 141.96 grams per mole) = 0.176 grams of Na2HPO4

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