When 1914.8C of electricity was passed through two electrolytic cells connected in series 0.6g of a metal M1 was deposited in one of the electrolyte, if 2.14g of another metal M2 was deposited at the other cathode and the sum of the quantity required to deposit 1mole of M1 and 1 mole of M2 is 289500C. Determine the charges on the ions of M1 and M2.
3 answers
50.0 mL of a 0.200 M solution of the weak base N-ethylmorpholine (C6H13NO) is mixed with 8.00 mL of 1.00 M HCl and then diluted to a final volume of 100.0 mL with water, the result is a buffer with a pH of 7.00. Compute the Kb of N-ethylmorpholine.
mmols weak base = 50.0 x 0.200 = 10
mmols HCl = 8.00 x 1.00 = 8
salt formed (the acid) = 8 mmols and concn = mmols/mL volume total.
weak base remaining = 2 mmols and concn = mmols/total volume
pH = pKa + log (base/acid). Substitute, solve for pKa, then
pKa + pK = pKw = 14 and solve for pKb and convert to Kb.
mmols HCl = 8.00 x 1.00 = 8
salt formed (the acid) = 8 mmols and concn = mmols/mL volume total.
weak base remaining = 2 mmols and concn = mmols/total volume
pH = pKa + log (base/acid). Substitute, solve for pKa, then
pKa + pK = pKw = 14 and solve for pKb and convert to Kb.
Note the typo in pKb.
mmols weak base = 50.0 x 0.200 = 10
mmols HCl = 8.00 x 1.00 = 8
salt formed (the acid) = 8 mmols and concn = mmols/mL volume total.
weak base remaining = 2 mmols and concn = mmols/total volume
pH = pKa + log (base/acid). Substitute, solve for pKa, then
pKa + pKb = pKw = 14 and solve for pKb and convert to Kb.
mmols weak base = 50.0 x 0.200 = 10
mmols HCl = 8.00 x 1.00 = 8
salt formed (the acid) = 8 mmols and concn = mmols/mL volume total.
weak base remaining = 2 mmols and concn = mmols/total volume
pH = pKa + log (base/acid). Substitute, solve for pKa, then
pKa + pKb = pKw = 14 and solve for pKb and convert to Kb.
1 answer