Im stuck on this question, can someone explain to me how to do it?
what is the pH for the solution made from 11.10mL of 0.2780 M NaOH solution mixed with 70.20mL of 0.1300M ammonium chloride solution? pKa for NH4+ = 9.241
a) 9.53
b) 9.24
c) 8.95
d) 9.71
e) 8.77
what is the pH for the solution made from 11.10mL of 0.2780 M NaOH solution mixed with 70.20mL of 0.1300M ammonium chloride solution? pKa for NH4+ = 9.241
a) 9.53
b) 9.24
c) 8.95
d) 9.71
e) 8.77
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Well I'm guessing this is a buffer system so you use the henderson equation:
pH = pKa + log(base/acid)
instead of the conc. of base and acid you number or moles because after they are mixed their concentrations change.
NaOH => n = CV = 11.1*.278 = 3.0858
NH4Cl => n = CV = 70.2*.13 = 9.126
(if you've noticed I didn't bother changing to L because you're going to divide them anyway so the units cancel and if you did change to L that would cancel)
pH = 9.241 + log ( 3.0858/9.216) = 8.77
pH = pKa + log(base/acid)
instead of the conc. of base and acid you number or moles because after they are mixed their concentrations change.
NaOH => n = CV = 11.1*.278 = 3.0858
NH4Cl => n = CV = 70.2*.13 = 9.126
(if you've noticed I didn't bother changing to L because you're going to divide them anyway so the units cancel and if you did change to L that would cancel)
pH = 9.241 + log ( 3.0858/9.216) = 8.77