2 added 18 characters in body edited Nov 14 '16 at 9:30 Satwik Pasani 4,18444 gold badges3636 silver badges6666 bronze badges I guess the mistake you are doing is in the equation for $$K_a$$ for $$\ce{CH3COOH}$$. See if the following makes sense. $$\frac{[\ce{H+}]^2}{[\ce{CH3COOH}]}=K_a(\ce{CH3COOH})$$$$\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}=K_a(\ce{CH3COOH})$$ Let $$x$$ be the $$[\ce{H+}]$$ from acetic acid, and $$y$$ from the other organic acid. $$\frac{10^{-2}}{0.1-x}=10^{-5}$$$$\frac{10^{-1}x}{0.1-x}=10^{-5}$$ $$0.09+x+y=0.1 \text{ (Balancing for H+)}$$ $$\frac{y^2}{[\ce{CHCl2COOH}]}=K_a(\ce{CHCl2COOH})$$$$\frac{10^{-1}y}{[\ce{CHCl2COOH}]}=K_a(\ce{CHCl2COOH})$$ This should give you the answer. I guess the mistake you are doing is in the equation for $$K_a$$ for $$\ce{CH3COOH}$$. See if the following makes sense. $$\frac{[\ce{H+}]^2}{[\ce{CH3COOH}]}=K_a(\ce{CH3COOH})$$ Let $$x$$ be the $$[\ce{H+}]$$ from acetic acid, and $$y$$ from the other organic acid. $$\frac{10^{-2}}{0.1-x}=10^{-5}$$ $$0.09+x+y=0.1 \text{ (Balancing for H+)}$$ $$\frac{y^2}{[\ce{CHCl2COOH}]}=K_a(\ce{CHCl2COOH})$$ This should give you the answer. I guess the mistake you are doing is in the equation for $$K_a$$ for $$\ce{CH3COOH}$$. See if the following makes sense. $$\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}=K_a(\ce{CH3COOH})$$ Let $$x$$ be the $$[\ce{H+}]$$ from acetic acid, and $$y$$ from the other organic acid. $$\frac{10^{-1}x}{0.1-x}=10^{-5}$$ $$0.09+x+y=0.1 \text{ (Balancing for H+)}$$ $$\frac{10^{-1}y}{[\ce{CHCl2COOH}]}=K_a(\ce{CHCl2COOH})$$ This should give you the answer. 1 answered Nov 14 '16 at 9:05 Satwik Pasani 4,18444 gold badges3636 silver badges6666 bronze badges I guess the mistake you are doing is in the equation for $$K_a$$ for $$\ce{CH3COOH}$$. See if the following makes sense. $$\frac{[\ce{H+}]^2}{[\ce{CH3COOH}]}=K_a(\ce{CH3COOH})$$ Let $$x$$ be the $$[\ce{H+}]$$ from acetic acid, and $$y$$ from the other organic acid. $$\frac{10^{-2}}{0.1-x}=10^{-5}$$ $$0.09+x+y=0.1 \text{ (Balancing for H+)}$$ $$\frac{y^2}{[\ce{CHCl2COOH}]}=K_a(\ce{CHCl2COOH})$$ This should give you the answer.