Answer
See answer below.
Work Step by Step
a) $pH=-log(0.105)=0.979$
b) $[H_3O^+]=10^{-2.56}=2.75\ mmol$, acidic
c) $[H_3O^+]=10^{-9.67}=2.14\times10^{-10}\ mol$, basic
d)$[H_3O^+]=(10.0\ mL\times2.56\ M)/250.\ mL=0.102\ M$
$pH=-log(0.102)=0.990$