91Ó°ÊÓ

To answer this problem, we can apply the Henderson-Hasselbach equation, which is as follows:

\(HF(aq) \to {H^ + }(aq) + {F^ - }(aq)\)

For this reaction, \({K_a}\)is determined as follows:

\({K_a} = \frac{{c\left( {{F^ - }} \right) \times c\left( {{H^ + }} \right)}}{{c(HF)}}\)

There's also a salt called \(NaF\)that dissociates in the water.

\(NaF(aq) \to N{a^ + }(aq) + {F^ - }(aq)\)

Now we'll make two assumptions that are common when working with buffers (this is a buffer because we have a weak acid and its salt).

Salt dissociation is the source of all \({\rm{N}}{{\rm{a}}^ + }\)in the \({K_a}\)equation.

\(c\left( {N{a^ + }} \right) = c(NaF)\)

We also assume that the \({\bf{HF}}\)equilibrium concentration in the \({K_a}\)equation is the same as its beginning concentration in the task:

\(c{(HF)_{eq}} = c{(HF)_{initial}}\)

We can now write \({K_a}\)as follows:

\({K_a} = \frac{{c(NaF) \times c\left( {{H^ + }} \right)}}{{c(HF)}}\)

We need to figure out \(c(NaF)\)

\(\begin{align}{K_a} \times c(HF) &= c(NaF) \times c\left( {{H^ + }} \right)\\c(NaF) &= \frac{{{K_a} \times c(HF)}}{{c\left( {{H^ + }} \right)}}\end{align}\)

The \({K_a}\)value for \({\bf{HF}}\)is \(3.5 \times {10^{ - 4}}\), according to the data in Appendix H of the book. We now have all of the information.:

\(\begin{align}c(NaF) &= \frac{{3.5 \cdot {{10}^{ - 4}} \times 0.300}}{{2.3 \times {{10}^{ - 4}}}}\\c(NaF) &= 0.457M\end{align}\)