Homework Answers
Number of moles of NaH2PO4=mass/molar mass
=14.0 g/120.0 g/mol
Concentration of NaH2PO4=number of moles/volume of solution (L)
=(14.0 g/120.0 g/mol)/0.940 L
Number of moles of Na2HPO4=mass/molar mass
=7.84 g/142.0 g/mol
Concentration of Na2HPO4=number of moles/volume of solution (L)
=(7.84 g/142.0 g)/0.940 L
Ka of H2PO4-=6.2 x 10-8
pKa=7.21
As per the henderson-hasselbalch equation pH of an acidic buffer is given as
pH=pKa+log[conjugate base]/[weak acid]
Where pKa=-logKa (Ka=acid dissociation constant)
=7.21 in this case
[Conjugate base]=concentration of conjugate base
=Concentration of Na2HPO4 in this case
[Weak acid]=Concentration of weak acid
=Concentration of NaH2PO4 in this case
Substitute the values in henderson-hasselbalch equation
pH=7.21+
=7.21 + 
=7.21+log(0.47)
=7.21-0.33=6.88
So pH of the resulting buffer=6.88 which is the last option
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What is the pH of the buffer that results when 14.0 g of NaH2PO4 and 7.84...
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If you need Ka values, you
should use this table:
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