Calculate the pH of a $3.0\%$ solution of sodium fluoride $(\rho_{solution}=1.030\frac{g}{mL})$
Find the species that influences the pH !
$F^-$ is a weak base
What is the initial molarity of this species ?
The number of moles of the fluoride ion is the same as that of sodium fluoride, since this salt dissociates entirely in water: Let us take $ 1 \; L $ of solution. $m_{solution}$ $=$ $1000\cdot 1.030$ $=$ $1030\;g$ $m_{NaF}$ $=$ $\frac{3.0\cdot 1030}{100}$ $=$ $30.9\;g$ $n_{F^-}$ $=$ $n_{NaF}$ $=$ $\frac{30.9}{42.0}=0.736\;mol$ $[F^-]=0.736\;M$
Equation ?
Taking $x=[OH^-]$, we have: $x^2\;+\;10^{-11.83}\;x\;-\;10^{-11.83}\;0.736\;=\;0$ $x=1.04\cdot 10^{-6}$
pH ?
$pOH\;=\;-log\; x\;=\;5.98$ $pH\;=\;14.00-5.98\;=\;8.02$
