The first step to answer this question is to state the dissociation equation for the given NaF:
[tex]\begin{gathered} NaF_\rightarrow Na_^++F_^- \\ F^-+H_2O\rightarrow HF+OH^- \end{gathered}[/tex]Now, we have to make an ICE table for the second reaction:
Find Kb using Ka, this way:
[tex]\begin{gathered} Ka\cdot Kb=1.00\times10^{-14} \\ Kb=\frac{1.00\times10^{-14}}{Ka} \\ Kb=\frac{1.00\times10^{-14}}{6.80\times10^{-4}} \\ Kb=1.47\times10^{-11} \end{gathered}[/tex]Use the expression that represents Kb:
[tex]Kb=\frac{\lbrack HF\rbrack\lbrack OH^-\rbrack}{\lbrack F^-\rbrack}[/tex]Replace for the value of Kb and the expressions for each concentration taken from the ICE table:
[tex]\begin{gathered} 1.47\times10^{-11}=\frac{x\cdot x}{3.06\times10^{-3}-x} \\ 1.47\times10^{-11}\cdot3.06\times10^{-3}=x^2 \\ x^2=4.50\times10^{-14} \\ x=2.12\times10^{-7} \end{gathered}[/tex]Calculate the pOH using the concentration of OH ions:
[tex]\begin{gathered} pOH=-log\lbrack OH^-\rbrack \\ pOH=-log\lbrack2.12\times10^{-7}\rbrack \\ pOH=6.67 \end{gathered}[/tex]Now, use the pOH to find the pH:
[tex]\begin{gathered} pH+pOH=14 \\ pH=14-pOH \\ pH=14-6.67 \\ pH=7.33 \end{gathered}[/tex]It means that the pH of the solution is 7.33.
