Question #fede8

2 Answers
Nov 9, 2017

Answer:

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#

Nov 9, 2017

Answer:

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)#