1.

How many grams of sodium acetate are to be added to a litre of 0.4 M CH_3 COOH solution so that the [H^+] of the resultant solution is 2 xx10^(-4)g ion/L? (Given K_a for acetic acid = 1.8 xx 10^(-5) )

Answer»

`3.502 g`
`2.952 g`
`4.252 g `
`5.162g `

Solution :`pH =pK_a+ log (["SALT"])/(["base"])`
`-log( 2xx 10^(-4) ) =- log( 1.8 XX 10^(-5) ) + log(["salt"])/( 0.4)`
`3.699 = 4.74+ log(["salt"])/(0.4 )`
` thereforelog(["salt"])/( 0.4 ) =- 1.041 or(["salt "])/(0.4 ) =0.091`
or `["salt "] = 3.6 xx 10^(-2)`
` therefore `AMOUNTOF sodiumacetateto beadded`= 3.6 xx 10^(-2)xx 82 = 2.952`g ( molarmass of`CH_3 COONa=82`)


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