Question

Explain what chemistry occurs in each region of the titration of OH with H

State how you would calculate the pH in each region.

Answer 1

OH- + H+ = H2O

there is neutralization

recall that

OH- = basic media, H = acidic media

we can always relate all of concentrations/Kw constants:

In order to be able to calculate pH, pOH and [H3O+]/[H+] and [OH-] we must understand the relationship between each of them

Remember that water will ionize as follows:

H2O(l) + H2O(l) <-> H3O+(aq) + OH-(aq)

so

Kw = [H3O+][OH-]

at T = 25°C, the Kw value is given as 10^-14

then,

10^-14 = [H3O+][OH-]

can be used always to relate H3O+ and OH-

also, note that

pH = -log([H+])

pOH = -log([OH-])

and

pH + pOH = 14 (stated above)

Therefore...

as we titrate OH- with H+

We keep adding H+ to OH-

each time some amount of OH- + H+ form H2O

this is called neutralization of an acid/base

the equivalence point:

the point in which OH- = H+

the endpoint --> the assumed value of volume in which we approximate OH- = H+

if this is a strong acid / strong base:

pH before equivalnece point

pH = 14- pOH = 14 - - log(OH) = 14 + log([OH-])

simply, get [OH-] left from H2O formation

[OH-] = mmol of OH- lef t/ Vtotal (including acid + bases)

in equivalence point

pH = 7, since [H+] = [OH-]

after extra acid is added

pH = -log([H+])

[H+] = mmol of H+ left after reaction / Vtotal (Vacid + Vabse)