10 mLof H,A (weak diprotic acid) solutions is titrated against 0.1M NaOH. pH of the solution is plotted against volume of strong base added and following observation is made. If pH of the solution at 1^(st) equivalence point is pH_(1) and at 2^(nd) equivalnee point is pH_(2), Cal the value (pH_(2),-pH_(1)) at 25^(@)C. Given for H_(2),A, p^(Kal) =4.6 &p^(Ka2) =8

10 mLof H,A (weak diprotic acid) solutions is titrated against 0.1M Na

1.	10 mLof H,A (weak diprotic acid) solutions is titrated against 0.1M NaOH. pH of the solution is plotted against volume of strong base added and following observation is made. If pH of the solution at 1^(st) equivalence point is pH_(1) and at 2^(nd) equivalnee point is pH_(2), Cal the value (pH_(2),-pH_(1)) at 25^(@)C. Given for H_(2),A, p^(Kal) =4.6 &p^(Ka2) =8
Answer» <html><body><p><br/></p>Solution :` H_2A +NaOH to <a href="https://interviewquestions.tuteehub.com/tag/naha-2855912" style="font-weight:bold;" target="_blank" title="Click to know more about NAHA">NAHA</a> +H_2O " at " I^(<a href="https://interviewquestions.tuteehub.com/tag/st-632849" style="font-weight:bold;" target="_blank" title="Click to know more about ST">ST</a>) " <a href="https://interviewquestions.tuteehub.com/tag/end-971042" style="font-weight:bold;" target="_blank" title="Click to know more about END">END</a> point,"` <br/> ` pH = (p^(K_(a_1))+p^(K_a_2))/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=( 8+ 4.6)/( 2) = 6.3` <br/> ` N_1V_1 =N_2V_2 `<br/> ` N_1xx 10=0 .1 xx 20 rArr N_1 =0.2` <br/> at ` 2^(nd) `end point , `Na_2 A` salt will be formed <br/>`{:(NaHA+ ,NaOHto ,Na_2 A + H_2O), (2m " moles " ,2m " moles",0),(-,-, 2 m " moles "):}` <br/> ` [Na_2 A]=(2)/(50)=0.04 M`</body></html>