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A buffer consisting of H2PO4^- and HPO4^2-, helps control the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the pH of a soft drink in which the major buffer ingredients are 6.5g of NaH2PO4 and 8.0g of Na2HPO4 per 355 mL of solution.

2007-03-27 13:58:00 · 2 answers · asked by pinkpanthergirl 1 in Science & Mathematics Chemistry

2 answers

MM(NaH2PO4)= 120 g/mol
6.5/120=0.0542 moles
Concentration NaH2PO4 = 0.0542/0.355 = 0.152 M

MM(Na2HPO4)=142 g/mol
8.0/142=0.0563 moles
Concentration Na2HPO4 =0.0563/0.355=0.158 M

NaH2PO4 >>Na+ + H2PO4-

Na2HPO4>> 2Na+ + HPO4-

The reaction is
H2PO4- <>H PO4- + H+
initial concentration
0.152..... .....0.158 .. ...0

at equilibrium
0.152-x... ... 0.158+x ....x

K = 6.2 10^-8 = (0.158+x)((x)/0.152-x

x=6 10^-8 M = concentration H+ >>pH=7.22

But from the dissociation of H2O Kw=(H+)(OH-)

we have (H+)=10^-7. We can not neglet this concentration so

Total concentration H+ = 6 10^-8 + 10^-7=1.6 10^-7

pH= 6.79

2007-03-28 05:12:20 · answer #1 · answered by Anonymous · 0 0

6.79

2014-04-16 22:21:50 · answer #2 · answered by Enda 2 · 0 0

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