Asked by Anonymous
Whats the pH of the solution which can be prepared by mixing same volumes of 0.5% NH4Cl and 0.25% KOH ?
(K=1.8 * 10^-5)
M(NH4Cl) 53.5 g/mol
M(KOH)=56g/mol
(K=1.8 * 10^-5)
M(NH4Cl) 53.5 g/mol
M(KOH)=56g/mol
Answers
Answered by
DrBob222
0.25g KOH/56 = about 0.004mols KOH
0.50g NH4Cl/53.5 = about 0.009 mols but you need to do these more accurately.
............NH4^+ + OH^- ==> NH3 + H2O
I...........0.009.....0........0
add.................0.004
C..........-0.004..-0.004.....0.004
E...........0.005.....0.......0.004
pH = 9.75 + log (0.004/0.009
Solve for pH (but get better numbers first.)
0.50g NH4Cl/53.5 = about 0.009 mols but you need to do these more accurately.
............NH4^+ + OH^- ==> NH3 + H2O
I...........0.009.....0........0
add.................0.004
C..........-0.004..-0.004.....0.004
E...........0.005.....0.......0.004
pH = 9.75 + log (0.004/0.009
Solve for pH (but get better numbers first.)
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