30 cc of (M)/(3) HCl , 20 cc of (M)/(2) HNO_(3) and40 cc of (M)/(4) NaOH solutions are mixed and the resulting solution is

30 cc of (M)/(3) HCl , 20 cc of (M)/(2) HNO_(3) and40 cc of (M)/(4) Na

1.	30 cc of (M)/(3) HCl , 20 cc of (M)/(2) HNO_(3) and40 cc of (M)/(4) NaOH solutions are mixed and the resulting solution is
Answer» <html><body><p>2<br/>1<br/>3<br/>8</p>Solution :Total millimolar of `H^(+)=(30xx(1)/(3))+(20xx(1)/(2))` <br/> `=10+10=20` <br/> Total <a href="https://interviewquestions.tuteehub.com/tag/millimoles-1096825" style="font-weight:bold;" target="_blank" title="Click to know more about MILLIMOLES">MILLIMOLES</a> of `OH^(-)=40xx(1)/(4)=10` <br/> `:. H^(+)` <a href="https://interviewquestions.tuteehub.com/tag/ions-1051295" style="font-weight:bold;" target="_blank" title="Click to know more about IONS">IONS</a> left after neutralization = 10 <a href="https://interviewquestions.tuteehub.com/tag/millimole-2833492" style="font-weight:bold;" target="_blank" title="Click to know more about MILLIMOLE">MILLIMOLE</a> <br/> Volume of solution `=1 dm^(3) = 1000cc` <br/> Hence, molarityof `H^(+)` ions `=(10)/(1000)M= 10^(-2)M` <br/> `:. <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> =- log [H^(+)]=-log 10^(-2)=2`</body></html>