1.

Whatis the pHof 1.0xx10^(-5)molarsolutionof KOH ?

Answer»

Solution :`KOH_((AQ)) impliesK_((aq))^(+) +OH_((aq))^(-)`Onemoleof KOHwouldgiveone moleof `OH^(-)`ions .
`[OH^(-)]=1xx10^(-5) mol "LITRE"^(-1)`
`POH=- log_(10)[OH ^-]=-LOG _(10) [10^5]`
`=-(5xx log _(10) ^(10))=-(-5)=5 `
`PH +POH= 14 `
`pH= 14 -pOH =14-5=9`


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