Answer
The identity of this salt is: $NaF$.
Work Step by Step
1. All the possible salts produces a basic solution (which is indicated by the pH). Since $Na^+$ doesn't affect the pH of the solution, the pH will be determined by the anion.
In that list, all the anions are conjugate pairs of weak acids, therefore, they are weak bases.
Find the $K_b$ for the base that would produce this solution:
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$B^-(aq) + H_2O(l) \lt -- \gt HB(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $HB$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [HB] = 0 + x = x$
-$[B^-] = [B^-]_{initial} - x $
For approximation, we are going to consider $[B^-]_{initial} = [B^-]$
** Notice:
$0.100 mole(Salt) = 0.100 moles(B^-)$
$\frac{0.100mol(B^-)}{1.0L} = 0.100M (B^-)$
3. Calculate the hydroxide ion concentration:
pH + pOH = 14
8.07 + pOH = 14
pOH = 5.93
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 5.93}$
$[OH^-] = 1.2 \times 10^{- 6}$
Therefore : $x = 1.2\times 10^{- 6}$
4. Write the Kb equation, and find its value:
$Kb = \frac{[OH^-][HB]}{ [B^-]}$
$Kb = \frac{x^2}{[Initial B^-] - x}$
$Kb = \frac{( 1.2\times 10^{- 6})^2}{ 0.10- 1.2\times 10^{- 6}}$
$Kb = \frac{ 1.4\times 10^{- 12}}{ 0.10}$
$Kb = 1.4\times 10^{- 11}$
5. Since $HB$ is the conjugate acid of $B^-$ , we can calculate its $K_a$ by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.4\times 10^{- 11} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.4\times 10^{- 11}}$
$K_a = 7.1\times 10^{- 4}$
This value is approximately the $K_a$ value for $HF$. Therefore, this salt is $NaF$.
