50 ml of 0.05 M Na_(2)CO_(3) is titrated against 0.1 M HCl. On adding 40 ml of HCl, pHof the resulting solution will be [{:(H_(2)CO_(3):pK_(a_(1))=6.35,pK_(a_(2))=10.33),(log3=0.477,log2=0.30):}]

50 ml of 0.05 M Na_(2)CO_(3) is titrated against 0.1 M HCl. On adding

1.	50 ml of 0.05 M Na_(2)CO_(3) is titrated against 0.1 M HCl. On adding 40 ml of HCl, pHof the resulting solution will be [{:(H_(2)CO_(3):pK_(a_(1))=6.35,pK_(a_(2))=10.33),(log3=0.477,log2=0.30):}]
Answer» <html><body><p>`6.35`<br/>`6.526`<br/>`8.34`<br/>`6.173`</p>Solution :`CO_(3)^(2-) + <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(-) rArr HCO_(3)^(-)`<br/> Initial millies <a href="https://interviewquestions.tuteehub.com/tag/moles-1100459" style="font-weight:bold;" target="_blank" title="Click to know more about MOLES">MOLES</a> `50 xx 0.05 40 xx 0.1` <br/> Final <a href="https://interviewquestions.tuteehub.com/tag/milli-560824" style="font-weight:bold;" target="_blank" title="Click to know more about MILLI">MILLI</a> moles `1.5 , 2.5`<br/> `{:(,HCO_(3)^(-)+,H^(+),rarr,H_(2)CO_(3)),("Initial milli mles",2.5,1.5,,"__"),("Final milli moles",1,"___",,1.5):}` <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> = pK_(a_(1)) + "log"([HCO_(3)^(-)])/([H_(2)CO_(3)]) = 6.173`</body></html>