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A 340.0 −mL buffer solution is 0.140 M in HF and 0.140 M in NaF. What...

A 340.0 −mL buffer solution is 0.140 M in HF and 0.140 M in NaF. What mass of NaOH could this buffer neutralize before the pH rises above 4.00? And, If the same volume of the buffer was 0.370 M in HF and 0.370 M in NaF, what mass of NaOHcould be handled before the pH rises above 4.00? Please show all steps. Thank you

Expert Solution

Sol 1.

Let the moles of NaOH added be x

initial moles of HF = 0.140 × 340.0 / 1000 = 0.0476 mol

initial moles of NaF = 0.140 × 340.0 / 1000 = 0.0476 mol

After addition of moles of NaOH , moles of NaF increases and moles of HF decreases

Final moles of HF = 0.0476 - x

Final moles of NaF = 0.0476 + x

Also , pKa of HF = 3.2

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( Final moles of NaF / Final moles of HF )

4 = 3.2 + log ( (0.0476 + x ) / ( 0.0476 - x ) )

log ( (0.0476 + x ) / ( 0.0476 - x ) ) = 4 - 3.2 = 0.8

(0.0476 + x ) / ( 0.0476 - x ) = 10^0.8 = 6.3096

0.0476 + x = 0.3003 - 6.3096x

x = 0.2527 / 7.3096 = 0.0346 mol

As molar mass of NaOH = 40 g / mol

So , mass of NaOH = 0.0346 × 40 = 1.384 g

Sol 2.

Let the moles of NaOH added be x

initial moles of HF = 0.370 × 340.0 / 1000 = 0.1258 mol

initial moles of NaF = 0.370 × 340.0 / 1000 = 0.1258 mol

After addition of moles of NaOH , moles of NaF increases and moles of HF decreases

Final moles of HF = 0.1258 - x

Final moles of NaF = 0.1258 + x

Also , pKa of HF = 3.2

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( Final moles of NaF / Final moles of HF )

4 = 3.2 + log ( (0.1258 + x ) / ( 0.1258 - x ) )

log ( (0.1258 + x ) / ( 0.1258 - x ) ) = 4 - 3.2 = 0.8

(0.1258 + x ) / ( 0.1258 - x ) = 10^0.8 = 6.3096

0.1258 + x = 0.7937 - 6.3096x

x = 0.6679 / 7.3096 = 0.0914 mol

As molar mass of NaOH = 40 g / mol

So , mass of NaOH = 0.0914 × 40 = 3.656 g

queen_honey_blossom answered 2 years ago

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