Question

Calculate the final pOH of a solution created by combining 125 mL of 2.50 M hydrochloric acid and 350 mL of 2.00 M sodium hydroxide.

Answer 1

$POH= - log(OH-)$

HCl +NaOH =NaCl + H2O

125ml of 2.00M solution contain (125/1000)*2=0.25moles of NaOH

POH=-log(0.25)

POH= 6.0