Question

What is the percent ionization of a 0.0741 M solution of a weak acid if the weak acid ionization constant, ka, is 2.0 x 10at 25 °C? Select one: a. 0.038% b. 0.30 % c. 1.4% d. 0.52% e. 5.2%

Answer 1

e the weak acid to be HA cad weak t dissociate completely. there is always acids do not dissociate completely. an extend of ionisation - Now Tw+] [A] Now, HAP=H7+ A =) ka=f Initially 0.0741 0 0 [WAT to J (0.07412)? at equm 0.0741 0.0741a 0.07410 0.0741(1-0) As XLI (weak aan Provided, (1-01 'low dissocialim) I HA7 = 0.0741(M) sa 10.07412) (1-0) so, ka = 0.0741

on, ka=0.074122 a= ka Vo.0741 = 0.07412 = 0.0741 ka V0.0741 = √ ka no.0741 = v 2x10-610.0741 ГАs ka=2310 ata 30, [nt] V2010-60.0741 (M) w nico WA Now, pencent ionisation [HA] V2210-620.0741 -x100 0.0741 = 0.52% (optiond)
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