100 mL of each pH = 3 and pH = 5 solutions of a strong acid are mixed. What is the resultant pH ?

100 mL of each pH = 3 and pH = 5 solutions of a strong acid are mixed.

1.	100 mL of each pH = 3 and pH = 5 solutions of a strong acid are mixed. What is the resultant pH ?
Answer» <html><body><p></p>Solution : <a href="https://interviewquestions.tuteehub.com/tag/ph-1145128" style="font-weight:bold;" target="_blank" title="Click to know more about PH">PH</a> of first solution `= <a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>, [H^(+) ]=10^(-3)=N_1`<br/>pH of second solution `= 5, [H^(+) ]=10^(-5) = N_2`<br/> Final <a href="https://interviewquestions.tuteehub.com/tag/proton-1170983" style="font-weight:bold;" target="_blank" title="Click to know more about PROTON">PROTON</a> concentration is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> by the equation:`[H^+]=(V_1 N_1+V_2 N_2)/(V_1 +V_2)`<br/> Substituting the values` [H^(+)] =(100 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(-3) xx 10^(-5))/( 100+100 ) = 5.05 xx 10^(-4)` M <br/> pH of the resultant mixture `= -log (5.05 xx 10^(-4) ) = 4 - log 5.05 = 3.3`</body></html>