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#16.51 A 340.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF....

#16.51 A 340.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF.

What mass of NaOH could this buffer neutralize before the pH rises above 4.00? ANSWER: 0.98

If the same volume of the buffer was 0.340 M in HF and 0.340 M in NaF, what mass of NaOHcould be handled before the pH rises above 4.00?

Solutions

Expert Solution

1) Pka (HF)=3.14

Let the required mass of NaOH be x

Moles of NaOH=mass/molar mass=x/40

Using Henderson-hasselbach equation,

pH=pka+log([base]/[acid] ,[base]=concentration of base in the buffer,[acid]=conc of acid in the buffer

[base]=0.130 mol/L*0.340L=0.0442 moles

[acid]= 0.130 mol/L*0.340L=0.0442 moles

If it neutralizes x/40 moles of NaOH,then moles of acid used up =x/40

NaOH+HF=NaF+H2O

And an equal amount of base is formed.

[base]’=0.0442+x/40

[acid]’=0.0442-x/40

At pH=4.00,

4.00=3.14+ log (0.0442+x/40)/( 0.0442-x/40)

Or,4.00-3.14= log (0.0442+x/40)/( 0.0442-x/40)

Or,0.86= log (0.0442+x/40)/( 0.0442-x/40)

10^0.86= (0.0442+x/40)/( 0.0442-x/40)

Or,7.24= (0.0442+x/40)/( 0.0442-x/40)

Or,7.24(0.0442-x/40)=0.0442+x/40

Or,0.32-7.24x/40=0.0442+x/40

Or,0.276=8.24x/40

Or.11.04=8.24x

X=1.3 g

[base]=0.340mol/L*0.340L=0.1156 moles

[acid]= 0.340mol/L*0.340L=0.1156 moles

If it neutralizes x/40 moles of NaOH,then moles of acid used up =x/40

NaOH+HF=NaF+H2O

And an equal amount of base is formed.

[base]’=0.1156+x/40

[acid]’=0.1156-x/40

At pH=4.00,

4.00=3.14+log (0.1156+x/40)/( 0.1156-x/40) [Henderson-hasselbach eqn)

Or, Or,4.00-3.14= log (0.1156+x/40)/( 0.1156-x/40)

Or,0.86= log (0.1156+x/40)/( 0.1156-x/40)

10^0.86= (0.1156+x/40)/( 0.1156-x/40)

Or,7.24= (0.1156+x/40)/( 0.1156-x/40)

Or,7.24(0.1156-x/40)=0.1156+x/40

Or.0.837-7.24x/40=0.1156+x/40

Or,0.721=8.24x/40

Or,x=3.5 g

queen_honey_blossom answered 1 year ago

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