A weak base MOH was titrated against a strong acid. The `pH` at `1/4` the equivalence point was 9.3. What will be the `pH` at `3/4` th equivalence point in the same titration? `(log3=0.48)`

Answer

`8.34`

Explanation

`pOH` at `1/4` the equivalence point `=14-9.3=4.7`
At `1/4` the equivalence point
`pOH=pK_(b)+"log" (["Salt"])/(["Base"])`
`4.7=pK_(b)+"log" (0.25x)/(0.75x)`
`4.7=pK_(b)+"log"1/3`
`pK_(b)=4.7+log3`
At `3/4` th equivalence point
`pOH=pK_(b)+"log"(["Salt"])/(["Base"])`
`pOH=4.7+log3 +"log"(0.75x)/(0.25x)`
`pOH=4.7+2xxlog3`
`pOH=4.7+2xx0.48`
`pOH=5.66`
`=pH=14-5.66=8.34`