Answer
$3.13$
Work Step by Step
Concentration of the acid:
$[H_3O^+]=10^{-1.92}=0.012\ M$
Number of moles of hydronium remaining after the neutralization:
$0.25\ L\times 0.012\ M-0.0105\ M\times 0.250\ L=0.375\ mmol$
Concentration:
$0.375\ mmol/500\ mL=0.75\ mM$
pH: $-log(0.00075)=3.13$
