1.

Following solution are prepared by the mixing different volumes of NaOH of HCl different concentrations.

Answer»

`60 mL M/10 HCl+40 mL M/10 NaOH`
`55 mL M/10 HCl+45 mL M/10 NaOH`
`75 mL M/5 HCl+25 mL M/5 NaOH`
`100 mL M/10 HCl+100 mL M/10 NaOH`

SOLUTION :`75 mL M/5 HCl+25 mL M/5 NaOH`
No. of moles of HCl= `0.2 TIMES 75 times 10^-3=15 times 10^-3`
No of Moles of NaOH=`0.2 times 25 times 10^-3=5 times 10^-3`
No of moles of HCl after mixing `=15 times 10^-3 -5 times 10^-3`
`therefore"Concentration"= ("No. of moles of HCl")/("Vol in litre")=(10 times 10^-3)/(100 times 10^-3)=0.1M`
for (III) solution pH of 0.1 M HCl=`-log_10(0.1)=1`.


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