Calculate the ph for each of the following cases in the titration of 50.0 ml of 0.210 m hclo(aq) with 0.210 m koh(aq).

(a) before addition of any koh
(b) after addition of 25.0 ml of koh
(c) after addition of 35.0 ml of koh
(d) after addition of 50.0 ml of koh
(e) after addition of 60.0 ml of koh

(a) before addition of any KOH  = pH =-log [0.210]=0.68

(b) after addition of 25.0 mL of KOH  = pH =-log [0.210]=1.15

(c) after addition of 35.0 mL of KOH  = pH =-log [0.210]=1.43

(d) after addition of 50.0 mL of KOH  = pH =-log [0.210]=7.00

(e) after addition of 60.0 mL of KOH = pH =-log [0.210]=12.30

Explanation:

pH is calculated as follows:

pH = -log [H+], for acids

pOH = -log [OH-], for bases

pH = 14 - pOH

In this tritation, we add a base (KOH), to an acid (HClO), which neutralices each other in a stoiquiometric relationship of 1:1

So, reemplacing the values in each case we got:

(a) before addition of any KOH

: The concentration is equal to the concentration of HClO, beacuse no KOH has been added

pH =-log [0.210]=0.68

(b) after addition of 25.0 mL of KOH  

The concentration is equal to the amount of moles of HClO, that haven´t reacted with the moles of KOH cantained in the 25 ml added, divided by the total volumen (50 ml + 25 ml) = 75 ml

to find the amount of moles in the 25ml, and the 50 ml:

moles KOH = volumen L * Molarity = 0.025 ml * 0.210 M = 5.25*

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*

moles HClO - moles KOH = 1.05*-5.25*=5.25*

concentration = moles / volume = = 0.070M

pH =-log [0.070]=1.15

(c) after addition of 35.0 mL of KOH  

The concentration is equal to the amount of moles of HClO, that haven´t reacted with the moles of KOH cantained in the 35 ml added, divided by the total volumen (50 ml + 35 ml) = 85 ml

to find the amount of moles in the 35ml, and the 50 ml:

moles KOH = volumen L * Molarity = 0.035 ml * 0.210 M = 7.35*

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*

moles HClO - moles KOH = 1.05*-7.35*=3.15*

concentration = moles / volume = = 0.037M

pH =-log [0.037]=1.43

(d) after addition of 50.0 mL of KOH  

moles HCl = moles KOH

pH = 7

(e) after addition of 60.0 mL of KOH

The concentration is equal to the amount of moles of KOH, that haven´t reacted with the moles of HClO cantained in the initial 50 ml, because now the KOH is in excess, divided by the total volumen (50 ml + 60 ml) = 100 ml

to find the amount of moles in the 50ml, and the 60 ml:

moles KOH = volumen L * Molarity = 0.060 ml * 0.210 M = 1.26*

moles HClO = volumen L * Molarity = 0.050 ml * 0.210 M = 1.05*

moles KOH - moles HClO = 1.26*-1.05*=2.1*

concentration = moles / volume = = 0.020M

pOH =-log [0.020]=1.70

pH = 14 - pOH = 12.3

explanation:

two years ago, when picking flowers in the mountains back of yosemite valley, i found a book. it was blotted and storm-beaten; all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however; many of the inner pages were well preserved. and although all were more or less stained and torn, whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today; its granite pages have been torn and blurred by the same storms that wasted the castaway book.

two years ago, when picking flowers in the mountains back of yosemite valley; i found a book. it was blotted and storm-beaten, all of its outer pages were mealy and crumbly. the paper seemed to dissolve like the snow beneath which it had been buried, however, many of the inner pages were well preserved. and although all were more or less stained and torn; whole chapters were easily readable. this is the condition of the great open book of yosemite glaciers today, its granite pages have been torn and blurred by the same storms that wasted the castaway book.

the objects have the same charge.