An organic acid (RCO2H) has 0.090 M concentration. If its pH is 1.53 calculate the Ka?

Question

#RCO_2H (ac) + H_2O ⇌ RCO_2^- #
# + H_3O^+ (ac)#

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Answer 1 · 2018-04-25T05:12:28

#K_a=0.0144#

We address the equilibrium...

#RCO_2H(aq) + H_2O(l) rightleftharpoonsRCO_2^(-) +H_3O^+#

...for which #K_a=([RCO_2^(-)][H_3O^+])/([RCO_2H])#

But we gots #[H_3O^+]=10^(-1.53)=0.0295=[RCO_2^(-)]#

And .....

#[RCO_2H]=(0.090-0.0295)*mol*L^-1=0.0605*mol*L^-1#

#K_a=(0.0295)^2/(0.0605)=0.0144#