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A 360.0?mL buffer solution is 0.160M in HFand 0.160M in NaF. What mass of NaOH could...

A 360.0?mL buffer solution is 0.160M in HFand 0.160M in NaF.

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

If the same volume of the buffer was 0.370M in HF and 0.370M in NaF, what mass of NaOHcould be handled before the pH rises above 4.00?

Expert Solution

given 360 ml of 0.160 M NaF and 0.160 M HF

moles of NaF = molarity x volume /1000

moles of NaF = 0.16 x 360 /1000

moles of NaF = 0.0576

moles of HF = 0.0576

Let moles of NaOH be y .

NaOH + HF -----> NaF + H20

from the reaction we get

1 mole of NaOH reacts with 1 mole of HF to give 1 mole of NaF

so y mole of NaOH reacts with y mole of HF to give y mole of NaF

new moles of HF = 0.0576 - y

new moles of NaF = 0.0576 + y

according to hasselbach henderson equation

we get

pH = pKa + log [ salt /acid]

pH = pKa + log [ NaF / HF]

4 = 3.17 + log [ 0.0576 + y / 0.0576 - y ]

y = 0.04275

mass of NaOH = moles x molar mass

mass of NaOH = 0.04275 x 40

mass of NaOH = 1.71 g

so 1.71 g of NaOH is needed

given 360 ml of 0.370 M NaF and 0.370 M HF

moles of NaF = molarity x volume /1000

moles of NaF = 0.37 x 360 /1000

moles of NaF = 0.1332

moles of HF = 0.1332

Let moles of NaOH be y .

NaOH + HF -----> NaF + H20

from the reaction we get

1 mole of NaOH reacts with 1 mole of HF to give 1 mole of NaF

so y mole of NaOH reacts with y mole of HF to give y mole of NaF

new moles of HF = 0.1332 - y

new moles of NaF = 0.1332 + y

according to hasselbach henderson equation

we get

pH = pKa + log [ salt /acid]

pH = pKa + log [ NaF / HF]

4 = 3.17 + log [ 0.1332+ y / 0.1332- y ]

y = 0.0988

mass of NaOH = 0.098 x 40

mass of NaOH = 3.955

so 3.955 g of NaOH is needed

queen_honey_blossom answered 3 weeks ago

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