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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.

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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.