If the same volume of the buffer(340 mL) was 0.340 M in HF and 0.340 M in NaF, w

Question

If the same volume of the buffer(340 mL) was 0.340 M in HF and 0.340 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00?

I have attempted this problem multiple times and I have yet to obtain the correct answer. My answers have been: 3.5 g, 2.9 g, and 3.4 grams, but none of these are correct. Any help would be greatly appreciated.

If it helps, I received the correct answer for a very similar problem: 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? The answer I received was .98 g.

Explanation / Answer

a)

apply:

pH = pKa + log(NaF/HF)

pKa = 3.14

4.0 = 3.14 + log(NaF/HF)

initially

mmol of HF = MV = 0.34*340 = 115.6 mmol

mmol of NaF = MV = 0.34*340 = 115.6 mmol

after adding "x" mmol of NaOH, then, HF reacts and forms more NaF

mmol of HF = 115.6 - x

mmol of NaF = 15.6 + x

substitute in pH equation

4.0 = 3.14 + log(NaF/HF)

4.0 = 3.14 + log((15.6 + x)/(115.6 - x))

solve for "x" mmol of NAOh

10^(4-3.14) = (15.6 + x)/(115.6 - x)

7.24435 * (115.6 - x) = (115.6+ x)

7.24435 * (115.6) - 115.6 = x +7.24435 x

x = (7.24435 * (115.6) - 115.6 ) / (8.24435 ) = 87.556 mmol of NaOH

Mass of NaOH = 40 g

mass = mol/MW = 87.556*40 = 3502.24 mg of NaOH

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