Question

In: Chemistry

A 360.0 −mL buffer solution is 0.170 M in HF and 0.170 M in NaF. A)...

A 360.0 −mL buffer solution is 0.170 M in HF and 0.170 M in NaF.

A) What mass of NaOH could this buffer neutralize before the pH rises above 4.00?

B) If the same volume of the buffer was 0.360 M in HF and 0.360 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00? I keep getting 1.8 g and 3.5 g, which is wrong.

Expert Solution

The dissociation constant ,Ka of HF = 7.2 x 10^-4

Therefore pKa = -log Ka = 3.14

Now using Henersen Hassalbalch equation for buffers

pH = pKa + log [salt] / acid] pH = pKa + log [F-]/ [HF] = 3.14 + log 0.170/0.170 = 3.14, this is the initial pH of the given buffer
Now we need the buffer pH to reach maximum to pH = 4

So putting values in the equation
4.00 = 3.14 + log [F-]/ [HF]
4.00 - 3.14 = log [F-]/ [HF]
0.86 = log [F-]/ [HF]

taking antilog
7.24= log [F-]/ [HF]

On addition of base NaOH

NaOH + HF ---> NaF + H2O It shows that it will consume some moles of HF and produce equal moles of the salt

Let the moles of HF reacted = x

Intial moles of HF= initial moles of NaF = 0.170 x 0.360 = 0.0612
7.24 = 0.0612+x/0.0612-x
0.443 - 7.24 x = 0.0612+x
0.382 = 8.24 x
x = 0.046 = Moles of NaOH

mass NaOH = Moles X mol wt = 0.046 x 40 g/mol=1.84 grams

2) now again the intial pH = pKa [as concentration of salt = conc of acid]

therefore

So putting values in the equation
4.00 = 3.14 + log [F-]/ [HF]
4.00 - 3.14 = log [F-]/ [HF]
0.86 = log [F-]/ [HF]

taking antilog
7.24= log [F-]/ [HF]

On addition of base NaOH

NaOH + HF ---> NaF + H2O It shows that it will consume some moles of HF and produce equal moles of the salt

Let the moles of HF reacted = x
7.24 = 0.360+x/0.360-x
2.61 - 7.24 x = 0.360+x
2.25 = 8.24 x
x = 0.273 M

Moles of NaOH in 0.360 mL = 0.273 X 0.360 = 0.0928
mass NaOH = Moles X mol wt = 0.0928 x 40 g/mol= 3.93 g

queen_honey_blossom answered 1 year ago

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