JEE Main Chemistry question
A monoprotic acid in 0.1 M solution has Ka=1.0×10⁻⁵. The degree of dissociation of acid is:
Answer: (A)
Solution
α=√(Ka/C)=√(10⁻⁴)=0.01=1.0%.
A monoprotic acid in 0.1 M solution has Ka=1.0×10⁻⁵. The degree of dissociation of acid is:
α=√(Ka/C)=√(10⁻⁴)=0.01=1.0%.