# Question fede8

Nov 9, 2017

The pH is approximately $9.69$, I would grade this is fairly difficult.

#### Explanation:

Let's assume these salts totally dissolve in solution.

$K {H}_{2} P {O}_{4} r i g h t \le f t h a r p \infty n s {K}^{+} + {H}_{2} P {O}_{4}^{-}$
${K}_{2} H P {O}_{4} r i g h t \le f t h a r p \infty n s 2 {K}^{+} + H P {O}_{4}^{2 -}$

Starting with the first salt,

$2 g \cdot \frac{K {H}_{2} P {O}_{4}}{136 g} = 1.47 \cdot {10}^{-} 2 m o l$

and the second,

$4 g \cdot \frac{{K}_{2} H P {O}_{4}}{174 g} = 2.30 \cdot {10}^{-} 2 m o l$

We may assume the ions will be in equal moles.

Hence,

$\left[{H}_{2} P {O}_{4}^{-}\right] \approx 1.47 \cdot {10}^{-} 2 M$
where ${K}_{b} = 1.3 \cdot {10}^{-} 12$

$\left[H P {O}_{4}^{2 -}\right] \approx 2.30 \cdot {10}^{-} 2 M$
where ${K}_{{b}_{2}} = 1.6 \cdot {10}^{-} 7$

Water isn't ionizing, so we may ignore auto-ionization of water. We may organize the equilibrium expression favorably, like so,

$\sqrt{\left[{H}_{2} P {O}_{4}^{-}\right] \cdot {K}_{b}} = \left[O {H}^{-}\right] \approx 1.38 \cdot {10}^{-} 7 M$
$\sqrt{\left[H P {O}_{4}^{2 -}\right] \cdot {K}_{b}} = \left[O {H}^{-}\right] \approx 4.85 \cdot {10}^{-} 5 M$

From here, we may add these molarities, since there is $1 L$ of water to arrive at the total concentration of hydroxide ions.

$\therefore 4.86 \cdot {10}^{-} 5 M \approx \left[O {H}^{-}\right]$

And finally, we may obtain the pH!

$p H = 14 + \log \left[O {H}^{-}\right] \approx 9.69$

Nov 9, 2017

7.4

#### Explanation:

We are dealing with the 2nd ionisation of phosphoric(V) acid:

$\textsf{{H}_{2} P {O}_{4}^{-} r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + {H}^{+}}$

For which:

$\textsf{{K}_{a 2} = \frac{\left[H P {O}_{4}^{2 -}\right] \left[{H}^{+}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]} = 6.2 \times {10}^{- 8}}$

These are equilibrium concentrations.

Rearranging:

$\textsf{\left[{H}^{+}\right] = {K}_{a 2} \times \frac{\left[{H}_{2} P {O}_{4}^{-}\right]}{\left[H P {O}_{4}^{2 -}\right]}}$

It looks like we have a buffer recipe. Because sf(K_(a2)# is very small we can assume that the moles we are given in the question will approximate to the equilibrium moles.

$\therefore$$\textsf{{n}_{{H}_{2} P {O}_{4}^{-}} = \frac{2}{136} = 0.00147}$

and $\textsf{{n}_{H P {O}_{4}^{2 -}} = \frac{4}{174} = 0.0023}$

Since the total volume is common we can put these amounts straight into the equation for $\textsf{{H}^{+}} \Rightarrow$

$\textsf{\left[{H}^{+}\right] = 6.2 \times {10}^{- 8} \times \frac{0.00147}{0.0023} = 3.962 \times {10}^{- 8} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[3.962 \times {10}^{- 8}\right] = 7.4}$