15.8g of NaOH can this buffer neutralize before the pH rises above 4.00 and 36.8g of NaOH could be handled before the pH rises above 4.00.
Given that,
The volume of buffer solution [tex]=350mL[/tex]
The buffer solution in HF [tex]=0.150 m[/tex]
The buffer solution in NaF [tex]=0.150m[/tex]
The initial concentration of HF and NaF in millimoles can be calculated as,
The initial concentration of HF [tex]=350mL*\frac{0.15mmol HF}{1mL} =52.5mmol HF[/tex]
The initial concentration of NaF [tex]=52.5mmol NaF[/tex]
The initial concentration of [tex]F^{-} =52.5mmol F^{-}[/tex]
The basic neutralisation reaction is as follows,
[tex]OH^{-} _{aq} +HF_{aq}[/tex] → [tex]H_{2} O_{(l)} +F^{-1}[/tex]
Initial concentration x 52.5 52.5
Concentration Change [tex]-x[/tex] [tex]-x[/tex] [tex]+x[/tex]
Final concentration 0 [tex]52.5-x[/tex] [tex]52.5 + x[/tex]
By using Henderson Hasselbalch equation the value of x can be calculated as,
[tex]p^{H} =p^{Ka}+log\frac{[F^{-} ]}{[HF]} \\p^{H}-p^{Ka}=log\frac{[F^{-} ]}{[HF]}\\10^{(p^{H}-p^{Ka})} =log\frac{[F^{-} ]}{[HF]}\\p^{Ka} of HF = 7.2*10^{-4}[/tex]
[tex]10^{(4-(-log(7.2*10^{-4})) )} =\frac{52.5+x}{52.5-x} \\7.2=\frac{52.5+x}{52.5-x}\\7.2(52.5-x)=52.5+x\\378-7.2x=52.5+x\\x+7.2x=378-52.2\\8.2x=325.5\\x=39.6mmol[/tex]
∴ [tex]Mass of NaOH = 39.6mmol OH^{-} *\frac{40mg NaOH}{1mmol NaOH} *\frac{1g NaOH}{100mg NaOH} =15.8g[/tex]
The mass of NaOH could be handled before the pH rises above 4 for the same volume of the buffer were 0.350M in HF and 0.350M in NaF can be calculated as,
∴ Mass of NaoH [tex]= 15.8g*\frac{0.35mmol}{0.15mmol} =36.8g[/tex]
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