Chemstry acid- base titration

Chemistry
Tutor: None Selected Time limit: 1 Day

What is the pH at the equivalence point in the titration of 20.0 mL of 0.200 M HA with 0.200 M NaOH? The Ka for HA is 7.07×10−6 .

Feb 27th, 2015

here the number of equivalen of the acid is:

4*10^(-3), so we  have to require that the base has 4*10^(-3) number of equivalent, so the volume of the added base is:

(0.2/1000)*V = 4*10^(-3)

and then:

V = 20 mL

so the concentration of the resultant salt Cs is:

Cs= (4*10^(-3))/(40*10^(-3)) = 0.2 N

then the [OH-] concentration of our solution is:

[OH-] = sqrt(Kw Cs/Ka) = sqrt(10^(-14)*0.1/ (7.07*10^(-6)) = 1.19*10^(-5)

so the pOH = 4.92 and pH = 14- 4.92 = 9.08

please note that Kw is the ionic product of the water

Feb 27th, 2015

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