Question

Calculate the pH of a 3.06×10^-3 M solution of NaF, given that the Ka of HF = 6.80 x 10^-4 at 25°C.

Answer 1

The first step to answer this question is to state the dissociation equation for the given NaF:

`\begin{gathered} NaF_\rightarrow Na_^++F_^- \\ F^-+H_2O\rightarrow HF+OH^- \end{gathered}`

Now, we have to make an ICE table for the second reaction:

Find Kb using Ka, this way:

`\begin{gathered} Ka\cdot Kb=1.00*10^(-14) \\ Kb=(1.00*10^(-14))/(Ka) \\ Kb=(1.00*10^(-14))/(6.80*10^(-4)) \\ Kb=1.47*10^(-11) \end{gathered}`

Use the expression that represents Kb:

`Kb=(\lbrack HF\rbrack\lbrack OH^-\rbrack)/(\lbrack F^-\rbrack)`

Replace for the value of Kb and the expressions for each concentration taken from the ICE table:

`\begin{gathered} 1.47*10^(-11)=(x\cdot x)/(3.06*10^(-3)-x) \\ 1.47*10^(-11)\cdot3.06*10^(-3)=x^2 \\ x^2=4.50*10^(-14) \\ x=2.12*10^(-7) \end{gathered}`

Calculate the pOH using the concentration of OH ions:

`\begin{gathered} pOH=-log\lbrack OH^-\rbrack \\ pOH=-log\lbrack2.12*10^(-7)\rbrack \\ pOH=6.67 \end{gathered}`

Now, use the pOH to find the pH:

`\begin{gathered} pH+pOH=14 \\ pH=14-pOH \\ pH=14-6.67 \\ pH=7.33 \end{gathered}`

It means that the pH of the solution is 7.33.