Answer to Question #168232 in Inorganic Chemistry for Arya Prakash

Question #168232

Calculate the pH of 0.100moldm^-3 propanoic acid


1
Expert's answer
2021-03-03T02:00:34-0500

First we need to write the Ka expression

CH3CH2COOH(aq) "\\iff" CH3CH2COO- (aq) + H3O+(aq)



Ka "=" "\\dfrac{[H3O^+][CH3CH2COO^-]}{[CH3CH2COOH]}"


"Since [H3O^+] =[CH3CH2COO^-]"


Then Ka "=\\dfrac{[H3O^+]^2}{[CH3CH2COOH]}"


Now, Ka for Propanoic acid is 1.34x10-5 and concentration is 0.1M


Therefore, 1.34x10-5 "=" "\\dfrac{[H3O^+]^2}{0.1M}"


"[H3O^+]^2 = 1.34x10^-3 x 0.1M"


"[H3O^+] =" "\\sqrt{1.34x10^-3x0.1M}"


"[H3O^+] = 1.15758x10^-3"


"pH =-log(1.15758x10^-3)=2.94"


pH = 2.9



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