Question

# A 360.0 −mL buffer solution is 0.150 M in HF and 0.150 M in NaF. a)...

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

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