Step 1

$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{L}}\mathbf{}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{HF}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$** = 0.10 mol HF**

$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{L}}\mathbf{}\mathbf{\times}\frac{\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{F}}^{\mathbf{-}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{L}}}$** = 0.050 mol F ^{-}**

Step 2

(a) adding 0.050 mol of HCl → acid

mol F^{-} = 0

mol HF = 0.150 mol

A 1.0 L buffer solution is 0.10 M in HF and 0.050 M in NaF. Which action destroys
the buffer?

(a) adding 0.050 mol of HCl

(b) adding 0.050 mol of NaOH

(c) adding 0.050 mol of NaF

(d) none of the above

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