Question

Question #fede8

Answer 1

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

Explanation:

Let's assume these salts totally dissolve in solution.

`KH_2PO_4 rightleftharpoons K^(+)+H_2PO_4^(-)`
`K_2HPO_4 rightleftharpoons 2K^(+)+HPO_4^(2-)`

Starting with the first salt,

`2g * (KH_2PO_4)/(136g) = 1.47*10^-2mol`

and the second,

`4g * (K_2HPO_4)/(174g) = 2.30*10^-2mol`

We may assume the ions will be in equal moles.

Hence,

`[H_2PO_4^(-)] approx 1.47*10^-2M`
where `K_b = 1.3*10^-12`

`[HPO_4^(2-)] approx 2.30*10^-2M`
where `K_(b_2) = 1.6*10^-7`

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

`sqrt([H_2PO_4^(-)] * K_b) = [OH^-] approx 1.38*10^-7M`
`sqrt([HPO_4^(2-)] * K_b) = [OH^-] approx 4.85*10^-5M`

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

`therefore 4.86*10^-5M approx [OH^-]`

And finally, we may obtain the pH!

`pH = 14 + log[OH^-] approx 9.69`

Answer 2

7.4

Explanation:

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

`sf(H_2PO_4^(-)rightleftharpoonsHPO_4^(2-)+H^+)`

For which:

`sf(K_(a2)=([HPO_4^(2-)][H^+])/([H_2PO_4^-])=6.2xx10^(-8))`

These are equilibrium concentrations.

Rearranging:

`sf([H^+]=K_(a2)xx([H_2PO_4^(-)])/([HPO_4^(2-)]))`

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.

`:.` `sf(n_(H_2PO_4^-)=2/136=0.00147)`

and `sf(n_(HPO_4^(2-))=4/174=0.0023)`

Since the total volume is common we can put these amounts straight into the equation for `sf(H^+)rArr`

`sf([H^+]=6.2xx10^(-8)xx0.00147/0.0023=3.962xx10^(-8)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log[3.962xx10^(-8)]=7.4)`