1.

What is the PH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

Answer»

`2.0`
3
`7.0`
12.65

Solution :X ml of 0.1 m NAOH+xmL of 0.01MHCl
No. of moles of NaOH =`0.1 TIMES x times 10^-3=0.1x times 10^-3`
No. of moles of HCl= `0.01 times x times 10^-3=0.01x times 10^-3`
=`0.09x times 10^-3`
Concentration of NaOH = `(0.09x times 10^-3)/(2x times 10^-3)=0.045`
`[OH^-]=0.045`
`=2-log 4.5`
`=2-0.65=1.35`
`pH=14-1.35=12.65`


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