A monoprotic acid in a 0.1 M solution ionizes to 0.001 %. If its ionization constant is (10-x/100), the value of x is

Question

A monoprotic acid in a 0.1 M solution ionizes to 0.001 %. If its ionization constant is (10-x/100), the value of x is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For monoprotic acid, K a =( a 2 C /(1- a )) ≈ a 2 C where α is very small. α (0.001/100)=10-5 K a =10-5 × 10-5 × 0.1 =1 × 10-11 Since K a =(10- x /100)=10-11 10- x =100 × 10-11 Or x =9

Q. A monoprotic acid in a 0.1M solution ionizes to 0.001%. If its ionization constant is 10010−x​, the value of x is __

For monoprotic acid, Ka​=(1−a)a2C​≈a2C where α is very small. α1000.001​=10−5 Ka​=10−5×10−5×0.1 =1×10−11 Since Ka​=10010−x​=10−11 10−x=100×10−11 Or x=9

Q. A monoprotic acid in a $0.1 M$ solution ionizes to $0.001%$ . If its ionization constant is $\frac{1 0 ^{- x}}{100}$ , the value of $x$ is __