Question

In: Chemistry

A 360.0 ml buffer solution is 0.130 mol/L in HF and 0.130 mol/L in NaF.

Part A : what mass of NaOH could this buffer neutralize before the PH rises above 4.00?

Part B: If the same volume of the buffer was 0.370mol/L in NaF, what mass of NaOH could be handled before the pH rises above 4.0?

the Ka of HF is 3.5*10^-4 this is all the information given, I can look up more ka or kb values in my textbook if they are needed just comment but this should be sufficient.

Solutions

Expert Solution

we know that

moles = molarity x volume (L)

so initially

moles of HF = 0.13 x 360 x 10-3 =46.8 x 10-3

moles of NaF = 0.13 x 360 x 10-3 = 46.8 x 10-3

noow

for buufers

pH = -log Ka + log [NaF / HF]

4 = -log 3.5 x 10-4 + log [NaF / HF]

[NaF / HF] = 3.5

now

let y moles of NaOH be added

then

HF + OH ---> F- + H20

now after reaction

moles of HF = 46.8 x 10-3 - y

moles of NaF = 46.8 x 10-3 + y

now

[NaF / HF] = 3.5

46.8 x 10-3 + y / 46.8 x 10-3 - y = 3.5

y = 0.026

moles of NaOH = 0.026

mass = 0.026 x 40 = 1.04

1.04 grams of NaOH may be added

we know that

moles = molarity x volume (L)

so initially

moles of HF = 0.13 x 360 x 10-3 =46.8 x 10-3

moles of NaF = 0.37 x 360 x 10-3 = 133.2 x 10-3

noow

for buufers

pH = -log Ka + log [NaF / HF]

4 = -log 3.5 x 10-4 + log [NaF / HF]

[NaF / HF] = 3.5

now

let y moles of NaOH be added

then

HF + OH ---> F- + H20

now after reaction

moles of HF = 46.8 x 10-3 - y

moles of NaF = 133.2 x 10-3 + y

now

[NaF / HF] = 3.5

133.2 x 10-3 + y / 46.8 x 10-3 - y = 3.5

y = 6.8 x 10-3

moles of NaOH = 6.8 x 10-3

mass = 6.8 x 10-3 x 40 = 0.272

0.272 grams of NaOH may be added

queen_honey_blossom answered 1 year ago

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A​ 360.0 ml buffer solution is 0.130 mol/L in HF and 0.130 mol/L in NaF.

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A 360.0 ml buffer solution is 0.130 mol/L in HF and 0.130 mol/L in NaF.