Question

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A 25.0 mL buffer solution is 0.350 M in HF and 0.150 M in NaF. Calculate...

A 25.0 mL buffer solution is 0.350 M in HF and 0.150 M in NaF. Calculate the pH of the solution after the addition of 37.5 mL of .200 M HCl. The pKa for HF is 3.46.

Expert Solution

Solution-

Given:

Volume of buffer = 25.0 mL = 0.025 L

[HF] = 0.350 M

[ NaF] = 0.150 M

We know the equation

# mole of = molarity * Volume in L

# mole of HF = 0.350 M * 0.025 L = 0.00875 mole of HF

# mole of NaF = 0.150 M * 0.0025 L = 0.000375 ,ole of NaF

After addition of 37.5 mL (0.0375 L) of 0.200 M HCl

# mole of HCl = 0.200 M * 0.0375 L = 0.0075 mole of HCl

We use ICE chart

For HF:

HF (aq) + H2O( I ) ----> F^- (aq) + H3O ⁺ (aq)

I 0.00875 0 0

C -x + x +x

E ( 0.00875 -x) x x

Ka expression

Ka = [ F^- (aq)][ H3O ⁺ (aq)] / [ HF(aq) ]

Ka for HF = 6.8 x 10^-4

6.8 x 10^-4= x² / ( 0.00875 -x)

By using 5% approximation we neglect x at the denominator.

6.8 x 10^-4= X² / (0.0.00875)

X² = 6.8 x 10^-4 X 0.00875 = 0.0595 X 10^-4

X = 0.2439 X 10^-2

X = [H3O⁺] = 0.00243 mole of H3O

We add both moles of H3O from HCl and HF

Total moles = 0.0075 mole HCl + 0.00243 mol H3O+

=0.00993 mol H3O+

Now

PH = - log [H3O ⁺] = - log (0.00993) = 2.00

Answer pH of buffer solution after addition of HCl = 2.00

queen_honey_blossom answered 2 years ago

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