A 360.0 −mL buffer solution is 0.150 M in HF and 0.150 M in NaF.
a) What mass of NaOH can this buffer neutralize before the pH rises above 4.00? = 1.6
b)If the same volume of the buffer were 0.370 M in HF and 0.370 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00?
Sol.
(a) Let Conc. of NaOH added be x
When , NaOH added , then , conc. of F- increases and conc. of HF decreases .
So ,
Final Conc. of F- = Inital Conc. of F- + Conc. of NaOH added
= 0.150 + x
and , Final Conc. of HF = Initial Conc. of HF - Conc. of NaOH added = 0.150 - x
Also , pKa of HF = 3.17
So , Using Henderson - Hasselbalch equation ,
pH = pKa + log ( [F-] / [HF] )
4.00 = 3.17 + log ( (0.150 + x ) / ( 0.150 - x ) )
or , ( 0.150 + x ) / ( 0.150 - x ) = 100.83 = 6.76
0.150 + x = 6.76 ( 0.150 - x ) = 1.014 - 6.76 x
7.76 x = 0.864
x = 0.111
Therefore , Conc. of NaOH added = 0.111 M
As Volume of solution = 360.0 mL = 0.360 L
So , Moles of NaOH added = 0.111 × 0.360 = 0.03996 mol
As Molar Mass of NaOH = 40 g/mol
So , Mass of NaOH added = 0.03996 × 40 = 1.5984 g
(b)
Now , Final Conc. of F- = Inital Conc. of F- + Conc. of NaOH added = 0.370 + x
and , Final Conc. of HF = Initial Conc. of HF - Conc. of NaOH added = 0.370 - x
Also , pKa of HF = 3.17
So , Using Henderson - Hasselbalch equation ,
pH = pKa + log ( [F-] / [HF] )
4.00 = 3.17 + log ( (0.370 + x ) / ( 0.370 - x ) )
or , ( 0.370 + x ) / ( 0.370 - x ) = 100.83 = 6.76
0.370 + x = 6.76 ( 0.370 - x ) = 2.5012 - 6.76 x
7.76 x = 2.1312
x = 0.275
Therefore , Conc. of NaOH added = 0.275 M
As Volume of solution = 360.0 mL = 0.360 L
So , Moles of NaOH added = 0.275 × 0.360 = 0.099 mol
As Molar Mass of NaOH = 40 g/mol
So , Mass of NaOH added = 0.099 × 40 = 3.96 g
A 360.0 −mL buffer solution is 0.150 M in HF and 0.150 M in NaF. a)...
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