Answer to Question #109372 in General Chemistry for mya 17

Question #109372

If the Ka of a monoprotic weak acid is 2.9×10−6, what is the pH of a 0.31M solution of this acid?

Expert's answer

pH=-lg[H+];

[H+]=√(K_a*C(acid))=√0.31*2.9*10^-6=9.48*10^-4 ;

pH=-lg(9.48*10^-4)=3

Answer: pH=3

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