What difference is there in change of pH with volume of NaOH added near the equi

Question

Explanation / Answer

ph before mixing = -log[h3o+] NaOH ----->Na+ + OH- conc of naoh= [oh-] = 3.13M [h3o+ ]= Kw/ [oh-] (kw is ionic product of water=10^-14) [h30+] = 10^-14/ 3.13 = 0.32 x 10^-14 M ph= -log[ 0.32 x 10 ^-14] = -log0.32 - (-14)log10 = 0.49 +14=14.49 after mixing since there is water so naoh is diluted Vol of 3.13M naoh taken = 10 ml vol of water = 700 ml total volume of solution= 710 ml no neutralisation is there so, conc of NaOH in solution after mixing = 3.13/ 710= 0.0044 M h3o+] = kw/ [oh-] = 10^-14 / 0.0044 = 227.27x 10^-14M ph after mixing = -log[h3o+]= -log( 227.27 x 10^-14) =11.64 0.106 M NH+ (700ml) after mixing 10 ml of 3.13 M naoh = 10 x 3.13 /1000= 0.0313moles 700 ml of 0.106 M nh4+ = 700 x 0.106 /1000= 0.0742moles of nh4+ moles of nh4+ left unneutralized = 0.0742- 0.0313 =0.043moles vol of solution= 700+ 10 =710 ml SO MOLARITY of nh4+ in resulting solution=0.043x1000/710 =0.06M ph = -log(0.06) =1.22 0.106 M NH3 (700 ml)(base) after mixing 10 ml of 3.13 M of naoh = 10 x 3.13 /1000 = 0.0313 moles 7oo ml of 0.106 M nh3 = 700 x 0.106 /1000 =0.0742 moles total vol of solution =710 ml so, nh3 is a base it will increase ph oh- conc no neutralization is there molarity of solution = [oh-] = (0.0742 + 0.0313) x 1000/ 710 = 0.15 M [h3o+] =kw/ [oh-] = 10^-14/ 0.15= 6.67x 10^-14M ph = -log[ h3o+] = -log( 6.67x 10^-14) = 13.18 by adding buffer ph after mixing = ph before mixing= 14.49 bcz 0.106 moles of nh3 will neutralixe 0.106 mioles of nh4+ the net oh- ion conc will be due to naoh so ph will remain same

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.