A:
50 ml HAc 0.1M = 5mMol HAc (weak acid)
50 NaOH 0.1 M = 5 mMol NaOH (strong base)
Equivalence: 1:1, 5mMol:5 mMol => 50ml 0,.1 M NaOH
B:
#pH = -log sqrt(Ka * F)#
=> #pH = -log sqrt(1.8^-5 **0.1M)#
=> #pH = -log sqrt(1.8^6)#
=> #pH= -log 1.34 10^-3#
=> #pH = 2.87#
C:
#pH = pKa + log((eta OH^-)/(eta HA_"init" - eta OH^-))#
#pH = pKa + log(2.5 ^-3/(5.0^-3 - 2.5 ^-3))#
=>#pH = pKa + log(2.5^-3/2.5^-3)#
=>#pH = pKa + log1#
#log1# = 0, so:
pH=pKa = #-logKa = 4.74#
D:
#pH= 14 + log sqrt((C_a* C_b* K_w) / ((C_a + C_b)* Ka))#
#pH = 14 + log sqrt((0.1 * 0.1 * 1*10^-14) / ((0.1 + 0.1 )* 1.8*10^-5))#
#pH = 14 + log sqrt ((1*10^-16)/(0.2 * 1.8*10^-5))#
#pH = 14 + log sqrt ((1*10^-10)/(3.6)) = 14 -5.28#
#=> pH = 8.72#
