#"pH"# of #"HF"# given #["HF"] = "0.25 mol/L"#... Can I assume #["HF"] = ["H"_3"O"^(+)]# and calculate it even if it's a weak acid?

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Answer 1 · 2017-12-15T19:06:00

By no means; you need #K_a# for #HF(aq)#....

This site gives #pK_a(HF)=3.17# ...

And so we assess the equilibrium....

#HF(aq) + H_2O(l)rightleftharpoonsF^(-) + H_3O^+#

i.e. #K_a=10^(-3.17)=([H_3O^+][F^-])/([HF])#

Now initially, #[HF]=0.25*mol*L^-1#...we ASSUME that #[HF]">>"[H_3O^+]"/"[F^-]#...and so if we set #[H_3O^+]-=[F^-]-=x#

#K_a=10^(-3.17)=(x^2)/(0.25-x)#

#x_1~=sqrt(10^(-3.17)xx0.25)=0.0130*mol*L^-1#...and we can use this value for a second approx....

#x_2=sqrt(10^(-3.17)xx(0.25-0.0130))=0.0126*mol*L^-1#

And since our approximations are converging, we take successive approximations....

#x_3=sqrt(10^(-3.17)xx(0.25-0.0126))=0.0127*mol*L^-1#

And so #[H_3O^+]=[F^-]=0.0127*mol*L^-1#

#[HF]=(0.25-0.0127)*mol*L^-1=0.237*mol*L^-1#

#pH=-log_10(0.0127)=1.89#