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1 mL of 0.1 N HCl is added to 999 mL sol...

1 mL of 0.1 N HCl is added to 999 mL solution of NaCl. The pH of the resulting solution will be :

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The correct Answer is:
`10.85`
M mol of `H ^(+)` (initial ) `=200 xx 0.031 xx 2= 12.4`
M mol of `OH ^(-)` (initial) `= 84xx0.15 = 12.6`
M mol of `OH ^(-)` left after neutralization `= 0.2.`
`[OH^(-)] _("final")=(0.2)/(284) = 7xx 10^(-4) M, pOH =3.15 & pH = 10.85`
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