Calculate the number of mg of Mn2+ left unprecipitated in 100 mL of a 0.1000 M solution of MnSO4 to which enough Na2S has been added to make the final sulfide ion (S2-) concentration equal to 0.0394 M. Assume no change in volume due to the addition of Na2S. The pKsp of MnS is 14.960
2006-10-03 05:12:03 · 4 answers · asked by vem1225 1 in Science & Mathematics ➔ Chemistry
Calculate the number of mg of Mn2+ left unprecipitated in 100 mL of a 0.1000 M solution of MnSO4 to which enough Na2S has been added to make the final sulfide ion (S2-) concentration equal to 0.0394 M. Assume no change in volume due to the addition of Na2S. The pKsp of MnS is 14.960
2006-10-03 05:15:43 · answer #1 · answered by Albert 2 · 0⤊ 1⤋
-log([Mn2+][S2-]) = 14.960
or [Mn2+] = 1.0960*10^-15/0.0394 = 2.783x10^-14 M
Thus mass of Mn2+ left = 55 x 2.783x10^-14 x 0.1 x 1000 = 153.065x10^-12 mg
2006-10-03 05:43:22 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋
you are wasting time on line in answers intead of looking into the chapter of quantative analysis you are currently studying?? you cannot earn a degree like that.
2006-10-03 05:18:24 · answer #3 · answered by mr.phattphatt 5 · 0⤊ 0⤋
Oh my! Don't bother dear, take up sport instead it's much healthier.
2006-10-03 05:14:36 · answer #4 · answered by Anonymous · 0⤊ 1⤋