Homework Help

What is the pH if 10 mL of 0.3 M HCl is added to 100mL of a 0.13 M phosphate buffer...

user profile pic

bjackie | eNoter

Posted September 9, 2013 at 7:20 AM via web

dislike 1 like

What is the pH if 10 mL of 0.3 M HCl is added to 100mL of a 0.13 M phosphate buffer that has a starting pH of 7.2.

1 Answer | Add Yours

user profile pic

llltkl | College Teacher | Valedictorian

Posted September 9, 2013 at 8:23 AM (Answer #1)

dislike 1 like

Phosphate buffer consists of ions `H_2PO_4^-`  and `HPO_4^(2-)` in aqueous solution. The relevant pKa for phosphoric acid is 7.2, governing the following equilibrium:

`H_2PO_4^(-) stackrel rarr larr HPO_4^(2-) + H^+`

From the Henderson-Hasselbalch equation,

`pH=pK_a+log_(10)[HPO_4^(2-)]/[H_2PO_4^-]`

pK_a for `[H_2PO_4^-]` is 7.2.

For a buffer having pH=7.2,

`[HPO_4^(2-)]=[H_2PO_4^-]*10^(pH-pK_a)` `=[H_2PO_4^-]*10^(7.2-7.2)= [H_2PO_4^-]=0.13 M`

100 mL of buffer originally contained (100*0.13) mmol, i.e.  0.013 mols of both of these ions.

Addition of 10 mL 0.3 M HCl converts (0.3*10) mmol, i.e. 0.003 mol `HPO_4^(2-)` to `H_2PO_4^-` and at the same time, dilutes the total volume by 10 mL.

In the new situation,

`[HPO_4^(2-)]` =(0.013-0.003), i.e. 0.010 mol in 110 mL = 0.010*1000/110 M

`[H_2PO_4^-]` =(0.013+0.003), i.e. 0.016 mol= 0.016*1000/110 M

pH`=7.2+log((0.010*1000/110)/( 0.016*1000/110))`

`=7.2+log(10/16)`

`=6.99588`

`~~7.0`

(This amount of acid added to 100 mL pure water would have shown a pH of 1.56).

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes