Question

The ionisation constants of HF, HCOOH and HCN at 298K are 6·8 × 10^–4, 1·8 × 10^–4 and 4·8 × 10^–9 respectively. Calculate the ionisation constants of the corresponding conjugate bases.

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

We know,

straight K subscript straight a space cross times space straight K subscript straight b space equals space straight K subscript straight w space space space space or space space space straight K subscript straight b space equals space straight K subscript straight w over straight K subscript straight a For space HF space colon space straight K subscript straight a space equals space 6.8 space cross times space 10 to the power of negative 4 end exponent space space space space space space space space space space space left square bracket given right square bracket therefore space space space space space space space straight K subscript straight b space equals space fraction numerator 10 to the power of negative 14 end exponent over denominator 6.8 space cross times space 10 to the power of negative 4 end exponent end fraction space equals 1.5 space cross times space 10 to the power of negative 11 end exponent For space HCOOH space colon space straight K subscript straight a space equals space 1.8 space cross times space 10 to the power of negative 4 end exponent space space left square bracket given right square bracket

therefore space space space space space space straight K subscript straight b space equals space fraction numerator 10 to the power of negative 14 end exponent over denominator 1.8 space cross times space 10 to the power of negative 4 end exponent end fraction space space space space space space space space space space space space space space equals space 5.6 space cross times space 10 to the power of negative 11 end exponent For space HCN colon thin space straight K subscript straight a space equals space 4.8 space cross times space 10 to the power of negative 9 end exponent space space space space left square bracket given right square bracket therefore space space space space straight K subscript straight b space equals space fraction numerator 10 to the power of negative 14 end exponent over denominator 4.8 space cross times 10 to the power of negative 9 end exponent end fraction space equals space 2.08 space cross times space 10 to the power of negative 6 end exponent

For Daily Free Study Material Join wiredfaculty