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The pH of an aqueous solution of a 0.1 M solution of a weak monoprotic acid which is 1% ionised is :

Answer»

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Solution :`{:(HA,hArr,H^(+),+,A^(-)),(C(1-ALPHA),,c alpha,,c alpha):}`
`[H^(+)]= c alpha = 0.1xx(1)/(100)=10^(-3)`
`therefore PH = -log [H^(+)]=3`


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