Q1/ Weak acid has a Ka =3.8×10-5 (i)calculate the pH of 0.25 moldm-3 of the weak acid?

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Answer 1 · 2018-05-13T20:22:16

#pH=2.56#

We interrogate the acid base equilibrium...

#HA(aq)+H_2O(l) rightleftharpoonsH_3O^+ + A^(-)#

And so...#K_a=([H_3O^+][A^(-)])/([HA])=3.80xx10^-5#...

And so if #x*mol*L^-1# dissociate...

#3.0xx10^-5=x^2/(0.25-x)#

Given the SMALL #K_a#...we make the approx. that #0.25-x~=0.25#

And so #x_1=sqrt(3.0xx10^-5xx0.25)=2.74xx10^-3*mol*L^-1#

#x_2=2.72xx10^-3*mol*L^-1#

#x_3=2.72xx10^-3*mol*L^-1#

#pH=-log_10(2.72xx10^-3)=-(-2.56)=2.56...#